## Class 12 Physics Ray Optics And Optical Instruments Notes and Important Question

• September 24, 2020

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## Class 12 Physics Ray Optics Optical Instruments Notes

Introduction

• We will try to learn what is light and also will try to understand some important phenomenon related to light and applications of optics in different optical instruments like microscope, telescope etc.
• How the phenomenon of laws of reflection and refraction play important part in our day to day to life.
• Light helps us to see all the objects around us. The entire phenomenon related to this we try a have a look on them.
• Also try to understand why certain objects enhance properties of light more as compared to others.
• What is Light?
• Light is a form of energy which enables us to see things around us.
• Light travels in straight path.
• Speed of light is c =3 x 108m/s.
• Light waves are the most common form of Electromagnetic waves.
• Ray of light is the path of the light wave travelling from one point to another.
• Beam of light is bunch of rays of light.

Reflection of light by Spherical Mirrors

• Reflection of light is the bouncing back off a ray of light when it strikes a boundary between different media through which it cannot pass.
• Reflection helps us to see different objects in this world.
• Laws of Reflection :-
• The angle of reflection is equal to the angle of incidence (angle between incident ray and the normal). <i = <r
• The incident ray, reflected ray and the normal to the reflecting surface at the point of incidence all lie in the same plane.
• Note: – These laws are valid at each point on any reflecting surface whether plane or curved.

The incident ray, reflected ray and the normal to the reflecting surface lie in the same plane.

Spherical Mirrors

• A spherical mirror is a part of a reflective spherical surface and they are in sphere in shape.
• It is made up of a large number of extremely small plane mirrors.

Spherical Mirrors are of two types:-

1. Concave Mirror: –
• They are silvered on the inside of sphere.
• It is a converging mirror.
• In a Concave mirror when the rays of light reflect back they meet or converge at a point that is why known as converging mirror.
1. Convex Mirror: –
• They are silvered on the outside of sphere.
• In Convex mirror the rays don’t meet at appoint after reflection that is why it is a diverging mirror.

Some terminologies related to Spherical Mirrors:-

Pole:

• The centre of the reflecting surface of a spherical mirror.
• It lies on the surface of the mirror.
• The pole is usually represented by the letter P.
1. Centre of curvature:
• The reflecting surface of a spherical mirror forms a part of a sphere. The sphere’s centre is called as centre of curvature.
• It is represented by the letter C.
• The centre of curvature is not a part of the mirror. It lies outside its reflecting surface.
• The centre of curvature of a concave mirror lies in front of it.
• However, it lies behind the mirror in case of a convex mirror.
• The radius of the sphere of which the reflecting surface of a spherical mirror forms a part
• It is represented by the letter R.
1. Principal axis
• A straight line passing through the pole and the centre of curvature of a spherical mirror.
• Principal axis is normal to the mirror at its pole.
1. Principal Focus
• Rays parallel to the principal axis falling on a concave mirror meet/intersect at the point on the principal axis. The point is called principal focus of concave mirror.
• The reflected rays appear to come from a point on the principal axis when rays parallel to the principal axis fall on a convex mirror, that point is called principal focus of convex mirror.
• The principal focus is represented by the letter F.
• The distance between the pole and the principal focus of a spherical mirror is called the focal length. It is represented by the letter f.
1. Aperture
• The diameter of the reflecting surface of spherical mirror is called its aperture.
• Mirrors whose aperture is much smaller than its radius of curvature, we use R=2f.

Image formation by Spherical Mirrors

• The rays of light through centre of curvature retrace its path.
• The ray of light parallel to the principal axis, on reflection, passes through the focus.
• The ray of light through F is reflected parallel to the principal axis.
• There are 2 ways an image is formed: –
• Real image of an object where reflected rays actually meet, they are inverted and formed on the same side of the object.
• Virtual image of an object where reflected rays appear to meet. They are always erect and cannot be projected on the screen.
• They are formed beyond the mirror

Sign Convections:-

Following are the sign convections which are to be followed:

1. The pole (P) of the mirror is taken as the origin. The principal axis of the mirror is taken as the x-axis (X’X) of the coordinate system.

(2) The object is always placed to the left of the mirror. This implies that the light from the object falls on the mirror from the left-hand side.

(3) All distances parallel to the principal axis are measured from the pole of the mirror.

(4) All the distances measured to the right of the origin (along + x-axis) are taken as positive while those measured to the left of the origin (along – x-axis) are taken as negative.

(5) Distances measured perpendicular to and above the principal axis (along + y-axis) are taken as positive whereas which are measured along (-y axis) are taken as negative.

(vi) The heights measured upwards with respect to x-axis and normal to the principal axis (x-axis) of the mirror/ lens is taken as positive. The heights measured downwards are taken as negative.

(vii) The radius of curvature and the focal length of a concave mirror are negative and those for a convex are positive.

Mirror Formula

• Concave Mirror
• (1/v) + (1/u) = (1/f)
• Where:-
• The distance of the object from its pole is called the object distance (u).
• The distance of the image from the pole of the mirror is called the image distance (v).
• The distance of the principal focus from the pole is called the focal length (f).
• Mirror equation relates image distance (v) with object distance (u) and focal (f) length
• Consider a concave mirror, an object AB placed in the front of the concave mirror. The image produced is A’B’.
• Consider two right-angled similar triangles A’B’F and MPF.
• Therefore (B’A’/PM) = (B’F/FP)
• Or (B’A’/BA) = (B’F/FP) (because PM= AB) (1)
• Since <APB = <AP’B’, the right angles triangles are also similar.
• Therefore, (B’A’/BA) = (B’P’/BP) (2)
• Comparing equations (1) and (2):-
• (B’F/FP) = (B’P –FP)/(FP) = (PB’)/(PB)   equation (A)
• Considering the sign conventions:
• Light travels from the object to the mirror MPN. Hence this is taken as the positive direction.
• The object AB, image A′B′ as well as the focus F from the pole P, we have to travel opposite to the direction of incident light.
• Therefore all are taken as negative.
• This implies, B’P = -v , FP =-f and BP =-u
• Therefore equation(A) changes to:
• (-v –f)/(-f) = (-v/-u)
• Or (v-f)/(f) =(v/u)
• => (1/v) +(1/u) = (1/f)
• This relation is known as the mirror equation.

(Concave mirror)

(b) In case of convex mirror

• Using equation (A) (B’F/FP) = (B’P –FP)/(FP) = (PB’)/(PB)
• Image Distance PB’ = (+v) as it is measured from the pole and in the direction of the incident ray.
• PF= + f as it is also along the incident ray.
• Object distance PB= (-u) as it is in the direction opposite to the incident ray.
• Substituting the values,
• (v-f)/(u) = (+v/-u)
• =>(v/f) – 1 = ((-v/u)
• Dividing throughout by f,
• (1/f) – (1/v) =(-1/u)
• Or (1/f) = (1/v) –(1/u)
• Therefore the mirror equation will be:- (1/f) = (1/v) –(1/u)

Magnification of a Spherical Mirror

• Linear Magnification (m) produced by a spherical mirror gives the relative extent to which the image of an object is magnified with respect to the object size.
• Mathematically,
• m= (height of the image h’)/(height of the object h)
• Or, m = (h’)/(h)
• If the sign of magnification is negative then the image is real and if it is positive then the image is virtual.
• A real, inverted image is formed by a concave mirror.
• In triangles A′B′P and ABP, we have,
• (-B’A’/BA) =(-B’P)/(BP) (From equation 2)
• Applying sign convention, this becomes
• (– h’/h) = -(-v/- u)
• Therefore, m = (h’/h) = (- v/u)

Concave Mirror

• In case of convex mirror: Virtual and erect image is formed.
• m = (h’/h) = (A’B’)/(AB) = (PB’)/(PB)
• m= -(v/u)

Convex Mirror

Note: – It is valid for all the cases of reflection by a spherical mirror (concave or convex) whether the image formed is real or virtual.

Problem: ( Ray Optics And Optical Instruments )

A small candle, 2.5 cm in size is placed at 27 cm in front of a concave mirror of radius of curvature 36 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved?

Given:

Size of the candle, h =2.5cm

Image size = h’

Object distance, u = −27 cm

Radius of curvature of the concave mirror, R = −36 cm

Focal length of the concave mirror, f =(R/2) = (-36/2) =-18cm

By mirror formula,

(1/u) + (1/v) = (1/f)

Or (1/v) = (1/f) – (1/u)

= (1/-18) – (1/-27)

= (-3 +2)/ (54)

= (-1/54)

Therefore v = -54cm

Therefore, the screen should be placed 54 cm away from the mirror to obtain a sharp image.

The magnification of the image is given as:

m= (h’/h)

=-(v/u)

Therefore, h’ = (-v/u) x h

= – (-54/-27) x 2.5

=-5cm

The (-) ive sign shows the image is real and inverted. If the candle is moved closer to the mirror, the screen has to be moved away from the mirror to obtain the image on the screen.

However when the candle is moved to a distance less than 18cm from the mirror, the image will become virtual image and cannot be obtained on the screen.

Problem:- ( Ray Optics And Optical Instruments )

Suppose while sitting in a parked car, you notice a jogger approaching towards you in the side view mirror of R = 2 m. If the jogger is running at a speed of 5 m s–1, how fast the image of the jogger appear to move when the jogger is (a) 39 m, (b) 29 m, (c) 19 m, and (d) 9 m away.

From the mirror equation, we get, (1/f) = (1/u) + (1/v)

For convex mirror, since R = 2 m, f = 1 m. Then for u = –39 m,

v = (1/u) + (1/f)

=> v = (fu)/ (u-f)

Therefore putting f= +1m, u =-39m,

v= (39/40) m

Since the jogger moves at a constant speed of 5 m s–1, after 1 s the position of the image v (for u = –39 + 5 = –34m) is (34/35) m.

The shift in the position of image in 1 s is

= (39/40)-(34/35) = (1365 -1360)/ (1400)

= (5/1400) = (1/280) m

Therefore, the average speed of the image when the jogger is between 39 m and 34 m from the mirror, is (1/280) ms–1

Similarly, it can be seen that for u = –29 m, –19 m and –9 m, the speed with which the image appears to move is

(1/150) ms-1, (1/60) ms-1 and (1/10) ms-1 respectively

Although the jogger has been moving with a constant speed, the speed of his/her image appears to increase substantially as he/she moves closer to the mirror. This phenomenon can be noticed by any person sitting in a stationary car or a bus. In case of moving vehicles, a similar phenomenon could be observed if the vehicle in the rear is moving closer with a constant speed.

Refraction of Light

• Light does not travel in the same direction in all media .When a light ray passes obliquely from rarer medium (air) to a denser medium (water, glass) there is a change in its direction of propagation.
• This bending of light at the boundary when it passes from one medium to another is termed as refraction.
• Refraction is due to change in the speed of light as it enters from one transparent medium to another.
• Laws of refraction given by Snell’s law:
1. The incident ray, the refracted ray and the normal to the interface at the point of incidence, all lie in the same plane.
2. The ratio of the sine of the angle of incidence to the sine of angle of refraction is constant.
• The angles of incidence (i) and refraction (r) are the angles that the incident and its refracted ray make with the normal, respectively.
• n21 = (sin i)/(sin r)
• Where n21 = is a constant known as refractive index of the second medium with the first.
•
• Note: – The refractive index (n21) depends on the wavelength of light, also characteristic of pair of media.
• It is independent of the angle of incidence.
•
• If n21 >1, r<i then the refracted ray bends towards the normal. Medium 2 will be optically denser. So medium 2 is called optically denser medium as compared to medium 1.
• If n21 <1, r>i then the refracted ray bends away from the normal. Medium 2 will be optically denser. So medium 2 is called optically rarer medium as compared to medium 1.
3. Refraction through a parallel-side slab
• For a rectangular slab, refraction takes place at two interfaces (air-glass and glass-air).
• From the diagram it is clear that r2=i1e. emergent ray is parallel to the incident ray, there will be no deviation.
• But there will be lateral shift w.r.t the incident ray.
• Refractive index of medium 3 w.r.t medium 1 n31 =(n3/n1)
• Or n31 = (n3/n2) x (n2/n1) = (n32 x n21)
• Where
• n1 = refractive index of medium 1
• n2 = refractive index of medium 2
• n3 = refractive index of medium 3
• It can be written as: – (n31 x n21) where n31 is the refractive index of medium 3 w.r.t medium 1.

Refraction & Natural phenomena

• The refraction of light through the atmosphere is responsible for many interesting phenomena.
1. Advanced Sunrise and Delayed Sunset:-
• For example, the sun is visible a little before the actual sunrise and until a little after the actual sunset due to refraction of light through the atmosphere.
• By actual sunrise we mean the actual crossing of the horizon by the sun.
• The figure shows the actual and apparent positions of the sun with respect to the horizon. The refractive index of air with respect to vacuum is 1.00029.
• Due to this, the apparent shift in the direction of the sun is by about half a degree and the corresponding time difference between actual sunset and apparent sunset is about 2 minutes.
• The apparent flattening (oval shape) of the sun at sunset and sunrise is also due to the same phenomenon.
1. Twinkling of Stars:-
• All the stars have light of their own. Stars twinkle because by the time the light reaches our eye, it passes through the various layers of the atmosphere.
• At one point of time star appears to be at one position and at another minute it appears to be at another position.
• So if we see one object at two different places at a very frequent interval of time. Therefore we feel that the star is twinkling.
• This twinkling of star is also the phenomenon of refraction of light.

Total Internal Reflection

• When light travels from an optically denser medium to a rarer medium at the interface, it is partly reflected back into the same medium and partly refracted to the second medium. This reflection is called the internal reflection.
• In total internal reflection phenomenon there is no refraction and the entire incident ray will get reflected.
• Conditions for Total Internal Reflection are as follows:-
• Light ray travels from denser to rarer medium.
• When a ray of light moves from a denser medium to a rarer medium it moves away from the normal. The ray will get refracted at an r (angle of refraction).
• But if we keep increasing angle of incidence it will further move from normal and the angle of refraction will become less.
• A condition will come if we keep on increasing the angle of incidence, the angle of refraction will become equal to 900 and further increase in angle of incidence there will be no refraction will occur but reflection will take place.
• This is total internal reflection.
• The limiting factor beyond what total internal reflection will take place will be :-
• The angle of incidence should be greater than the angle of incidence for which angle of refraction is 900.
• Angle of incidence should correspond to angle of refraction = 900.
• This angle is known as Critical angle.
• Angle of incidence should be greater than Critical angle.
• Total internal Reflection is an optical phenomenon observed when a ray of light travelling from denser to rarer medium strikes the boundary at an angle greater than the critical angle.

Applications of Total Internal Reflection

• Optical Fibres:-
• They are used in telecommunication industries.
• Optical fibres work on the phenomenon of total internal reflection.
• Characteristics of Optical Fibres:-
• They are small in size and light in weight.
• They have greater information carrying capacities than metallic wires.

Working of Optical fibres:

• Optical fibres are fabricated with high quality composite glass/quartz fibres.
• Each fibre consists of a core and cladding. The refractive index of the material of the core is higher than that of the cladding.
• As there is difference in the refractive index of core and denser; core acts as a denser medium and cladding acts as a rarer medium.
• When a signal in the form of light is directed at one end of the fibre at a suitable angle, it undergoes repeated total internal reflections along the length of the fibre and finally comes out at the other end.
• Since light undergoes total internal reflection at each stage, there is no appreciable loss in the intensity of the light signal.
• Optical fibres are fabricated such that light reflected at one side of inner surface strikes the other at an angle larger than the critical angle.
• Even if the fibre is bent, light can easily travel along its length. Thus, an optical fibre can be used to act as an optical pipe.
• They are made up of plastic.

Glass vs. Plastic Optical Fibres

Applications of Optical Fibres

1. Fibre optic endoscopy
1. Decorative items
1. They are used in communication system
1. Prism
• Prisms make use of total internal reflection which makes it useful in binoculars.
• Prisms designed to bend light by 90º or by 180º make use of total internal reflection.
• Such a prism is also used to invert images without changing their size.
• In the first two cases, the critical angle ic for the material of the prism must be less than 45º.

Problem: ( Ray Optics And Optical Instruments )

A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism.

The angle of minimum deviation is measured to be 40°. What is the refractive index of the material of the prism?

The refracting angle of the prism is 60°. If the prism is placed in water (refractive index 1.33), predict the new angle of minimum deviation of a parallel beam of light.

Given:

Angle of minimum deviation, δm = 40°

Angle of the prism, A = 60°

Refractive index of water, µ = 1.33

Refractive index of the material of the prism = µ’

The angle of deviation is related to refractive index (µ’) as:

µ’= (sin (A + δm)/2)/ (sin (A/2))

= (sin (600 + 400))/ (sin (600/2))

= (sin (500))/ (sin (300))

= 1.532

Hence, the refractive index of the material of the prism is 1.532.

Since the prism is placed in water,

Let δm ‘= the new angle of minimum deviation for the same prism.

The refractive index of glass with respect to water is given by the relation:

μgw = (µ’)/ (µ)

= (sin (A + δm ‘)/ (2)) / (sin (A/2))

(sin (A + δm ‘)/ (2)) = (sin (A/2)) (µ’)/ (µ)

(sin (A + δm ‘)/ (2)) = sin (600) (1.532)/ (1.33)

(sin (A + δm ‘)/ (2)) = 0.5759

(A + δm ‘)/ (2) = sin -1(0.5759)

(A + δm ‘)/ (2) = 35.160

(600 + δm ‘) = 70.320

Therefore, δm ‘= (70.320 – 600)

δm ‘=10.320

Hence, the new minimum angle of deviation is 10.32°

Refraction at Spherical Surfaces

• Consider refraction at a spherical interface between two transparent media. An infinitesimal part of a spherical surface can be regarded as planar and the same laws of refraction can be applied at every point on the surface.
• The rays are incident from a medium of refractive index n1, to another of refractive index n2.
• Assuming the aperture (or the lateral size) of the surface to be small compared to other distances involved, so that small angle approximation can be made.
• Consider NM will be taken to be nearly equal to the length of the perpendicular from the point N on the principal axis.
• Considering small angles,
• tan<NOM = (MN)/(OM)
• tan < NCM = (MN/MC)
• tan<NIM = (MN)/(MI)
• Now, for ΔNOC, <i is the exterior angle. Therefore, i = ∠NOM + ∠NCM
• i = (MN/OM)+(MN/MC) (equation (1))
• Similarly, r = ∠NCM – ∠NIM
• e., r = (MN/MC) – (MN/MI) (equation (2))
• Now, by Snell’s law n1 sin i = n2 sin r or for small angles
• n1i = n2r
• Substituting i and r from Equation. (1) and (2), we get
• (n1/OM) + (n2/MI) = (n2 – n1)/MC) Equation (3)
• Here, OM, MI and MC represent magnitudes of distances. Applying the Cartesian sign convention,
• OM = –u, MI = +v, MC = +R
• Substituting these in Eq. (3), we get,
• (n2 –v) –(n1/u) = (n2 – n1)/(R)  Equation (4)
• Equation (4) gives us a relation between object and image distance in terms of refractive index of the medium and the radius of curvature of the curved spherical surface. It holds for any curved spherical surface.

Problem:

A small pin fixed on a table top is viewed from above from a distance of 50cm. By what distance would the pin appear to be raised if it is viewed from the same point through a 15cm thick glass slab held parallel to the table? Refractive index of glass = 1.5. Does the answer depend on the location of the slab?

Given:

Actual depth of the pin, d = 15 cm

Apparent depth of the pin = d’

Refractive index of glass, μ = 1.5

Ratio of actual depth to the apparent depth is equal to the refractive index of glass, i.e. μ = (d/d’)

Therefore, d’ = (d/ μ)

= (15/1.5) =10 cm

The distance at which the pin appears to be raised = (d’ –d)

= (15-10) = 5cm

For a small angle of incidence, this distance does not depend upon the location of the slab.

Lens Maker’s Formula

• The image formation can be seen in terms of two steps:
• The first refracting surface forms the image I1 of the object O.
• The image I1 acts as a virtual object for the second surface that forms the image at I.
• Applying equation,
• (n1 /OM) + (n2/MI) = (n2 – n1)/MC)
• to the first interface ABC, we get
• (n1/OB) + (n2/BI1) = (n2-n1)/BC1) (Equation 1)
• A similar procedure applied to the second interface ADC :(n2 /DI1) + (n1/DI) =(n2 – n1)/(DC2) (Equation 2)
• (In this case medium on the right side of ADC is n1 while on left is n2. Also DI1 is (-) as the distance is measured against the direction of incident light.)
• For a thin lens, BI1 = DI1.
• Adding equation(1) and equation (2)
• (n1/OB) + (n1/DI) = (n2-n1)(1/BC1 + 1/DC2) Equation (3)
• Suppose the object is at infinity, i.e. OB-> ∞ and DI =f,
• Therefore from the Equation(3),
• (n1/f)=(n2-n1)(1/BC1 + 1/DC2) (Equation 4)
• The point where image of an object placed at infinity is formed is called the focus F, of the lens and the distance f gives its focal length.
• A lens has two foci, F and F′, on either side of it.
• By the sign convention,
• BC1 = + R1,DC2 = –R2
• Therefore equation(4) can be written as,
• (1/f)=(n21 -1)((1/R1)- (1/R2)) Equation(5) (because n21=(n2/n1))
• Equation (5) is known as Lens Maker’s Formula.
• It is useful to design lenses of desired focal length using surfaces of suitable radii of curvature.
• Note that the formula is true for a concave lens also. In that case R1is negative, R2 positive and therefore, f is negative.
• Therefore from equation 3 and 4 we get,
• (n1/OB) + (n1/DI) =(n1/f)
• In the thin lens approximation, we can take B and D so close to the optical centre of the lens.
• Applying sign conventions, BO=-u and DI=+ v
• (1/v)- (1/u) = (1/f). It is thin lens formula.
• This formula is valid for both convex and concave lenses.
• The two foci, F and F′, of a double convex or concave lens are equidistant from the optical centre. The focus on the side of the (original) source of light is called the first focal point, whereas the other is called the second focal point.

Power of a Lens

• The power P of a lens is defined as the tangent of the angle by which it converges or diverges a beam of light falling at unit distant from the optical centre.
• tan δ = (h/f) if h=1
• tan δ = (1/f)
• Or δ = (1/f) for small value of δ .
• Therefore, P =(1/f) Where, p = power of the lens and f = focal length of the lens in meters.
• Power of a lens is a measure of the convergence or divergence produced by a lens.
• A lens of shorter focal length bends the incident light more, converging in case of convex and diverging in case of concave.
• The SI unit for power of a lens is dioptre (D):
1D = 1m–1.
• The power of a lens of focal length of 1 metre is one dioptre.
• Power of a lens is positive for a converging lens and negative for a diverging lens.

Combination of thin lenses in contact

• Consider two lenses A and B of focal length f1 and f2 placed in contact with each other.
• An object is placed at O beyond the focus of the first lens A on the common principal axis.
• The image of point object O will be at I1 formed by lens A, which act as a virtual object for second lens B producing the final image at I.
• As these lenses are thin, therefore a common optical centre is chosen. Let this common optical centre be P.
• Let Object distance for the first lens(A) PO =u
• Final Image distance PI =v
• Final Image distance PI1 = v1
• Considering lens A, Image I1 is produced
• Therefore, (1/v1) – (1/u1) = (1/f1) (Equation (1))
• Considering lens B, Final Image I is produced
• Therefore, (1/v) – (1/v1) = (1/f2) (Equation (2))
• Adding (Equation (1)) and (Equation (2))
• (1/v) – (1/u) = (1/f1) + (1/f2) (Equation (3))
• If we replace this combination by a single lens of focal length F so that image of O is formed at the same position I,
• Therefore, (1/v)-(1/u) = (1/F) (Equation (4))
• (1/F) = (1/f1) + (1/f2) (Equation (5))
• The derivation is valid for any number of thin lenses in contact. If several thin lenses of focal length f1, f2, f3… are in contact.
• The effective focal length of their combination is given by:
• (1/f) = (1/f1) + (1/f2) + (1/f3) + … (Equation (6))
• In terms of power, Eq. (6) can be written as:
• P = P1 + P2 + P3 + … (Equation (7)) where P is the net power of the lens combination.
• The sum in Equation (7) is an algebraic sum of individual powers, so some of the terms on the right side may be positive (for convex lenses) and some negative (for concave lenses).
• Combination of lenses helps

(a) To meet desired magnification

(b) It also enhances sharpness of the image.

(c) To make final image erect.

(d) To remove certain defects in the lens.

• Such a system of combination of lenses is commonly used in designing lenses for cameras, microscopes, telescopes and other optical instruments.

Refraction by Lens: Convex & Concave

1. A ray of light incident on the lens parallel to the principal axis after refraction passes through second principal axis.
2. A ray of light passing through first principal focus after refraction should move parallel to the principal axis.
3. A ray of light passing through the optical centre should go-undeviated after refraction.

Magnification of a lens

• It is the ratio of the size of the image to the size of the object.
• It is denoted as m =(h’/h) = (v/u)
• For a virtual and erect image is formed by a convex lens or by concave lens, m is positive and for real and inverted image m is negative.

Problem: ( Ray Optics And Optical Instruments )

An object of size 3.0cm is placed 14cm in front of a concave lens of focal length 21cm. Describe the image produced by the lens. What happens if the object is moved further away from the lens?

Given:

Size of the object, h1 = 3 cm

Object distance, u = −14 cm

Focal length of the concave lens, f = −21 cm

Image distance = v

According to the lens formula, we have the relation:

(1/v) – (1/u) = (1/f)

(1/v) = – (1/21) – (1/14)

= (-2-3)/ (42)

= (-5/42)

Therefore, v = – (42/5) =- 8.4cm

Hence, the image is formed on the other side of the lens, 8.4 cm away from it. The negative sign shows that the image is erect and virtual.

The magnification of the image is given as:

m = (h2)/ (h1) = (v/u)

Therefore h2 = ((-8.4/-14) x 3)

h2 = 1.8cm

Hence, the height of the image is 1.8 cm.

If the object is moved further away from the lens, then the virtual image will move toward the focus of the lens, but not beyond it.

The size of the image will decrease with the increase in the object distance.

Problem:- ( Ray Optics And Optical Instruments )

Double-convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be 20cm?

Refractive index of glass, μ = 1.55

Focal length of the double-convex lens, f = 20 cm

Radius of curvature of one face of the lens = R1

Radius of curvature of the other face of the lens = R2

Radius of curvature of the double-convex lens = R

Therefore, R1 = R and R2 = -R

The value of R can be calculated from Lens – Maker formula:

(1/f) = (μ – 1) [(1/ R1) – (1/ R2)]

(1/20) = (1.55 -1) [(1/R) + (1/R)]

(1/20) = 0.55 x (2/R)

Therefore R = (0.55 x 2 x20)

=22cm

Hence, the radius of curvature of the double-convex lens is 22 cm.

Problem:- ( Ray Optics And Optical Instruments )

The image of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall 3m away by means of a large convex lens.

What is the maximum possible focal length of the lens required for the purpose?

Given:

Distance between the object and the image, d = 3 m

Maximum focal length of the convex lens = fmax

For real images, the maximum focal length is given as:

fmax =(d/4) = (3/4)

=0.75m

Hence, for the required purpose, the maximum possible focal length of the convex lens is 0.75 m.

Refraction through prism

• Prism is a transparent optical material with flat polished surfaces that refract light.

Terminologies

• Incident Ray: – The ray which enters the prism.
• Refracted Ray: – The ray which comes out of prism inside the prism.
• Emergent Ray: – The ray which comes out of prism.
• Angle of deviation δ:-The angle which will tell how much the emergent ray deviated from the original incident ray.
• Angle of Prism: – The angle of the prism <A is known as angle of prism.
• The angles of incidence and refraction at the first face AB are i and r1, while the angle of incidence (from glass to air) at the second face AC is r2 and the angle of refraction or emergence (e).
• The angle between the emergent ray RS and the direction of the incident ray PQ is called the angle of deviation, δ.
• Consider the quadrilateral AQNR, in the figure two angles (at the vertices Q and R) are right angles.
• <A + <QNR = 1800 (As sum of other angles of a quadrilateral is 1800).
• From the triangle QNR,(r1 + r2 + ∠QNR) = 180º
• Comparing these two equations, we get

(r1 + r2 = A)  (Equation 1)

• The total deviation δ is the sum of deviations at the two faces,
• δ = (i – r1 ) + (e – r2 ) that is,
• δ = (i + e – A)  (Equation 2)
• Thus, the angle of deviation depends on the angle of incidence.
• Relation between angle of deviation and angle of incidence:-
• As the angle of incidence increases the angle of deviation keep on decreasing and when it reaches a point where angle of incidence is equal to angle of emergence, then the angle of deviation is minimum, and again it will start decreasing.
• This implies angle of deviation is minimum

δmin when angle of incidence <i = angle of emergence <e.

Angle of Deviation

• At the minimum deviation Dmin, the refracted ray inside the prism becomes parallel to its base.
• Therefore, we have δ = Dm, i = e which implies r1 = r2.
• From (Equation 1) 2r = A or r = (2/A)
• In the same way, (Equation 2) gives

Dm = (2i – A), or i = ((A + Dm)/2) (Equation 3)

• The refractive index of the prism is

n21 = (n1/n2)

=sin [(A + Dm)/2]/ (sin [A/2])

≈ ((A + Dm)/ (2))/ (A/2)

• The angles A and Dm can be measured experimentally. Equation (3) thus provides a method of determining refractive index of the material of the prism.
• For a small angle prism, i.e., a thin prism,

Dm, is also very small = (n21 -1) A

• It implies that, thin prisms do not deviate light much.

Dispersion of light through Prism

• Splitting of light into its constituent colours.
• The pattern of colors obtained is termed as spectrum.
• The red light bends the least, while the violet light bends the most.
• The phenomenon of splitting of light into its component colours is known as dispersion.

Causes of Dispersion

• Newton performed the experiment in order to explain the phenomenon of Dispersion.
• He took 2 prisms and arranged them in such a manner when a white light enters the first prism it will split the white light into its constituent colours.
• These constituent colours when were allowed to enter second prism which was kept in the inverted position.
• As a result again a white light was obtained.
• This experiment proved that it is the property of the prism because of which white light is getting split into its constituent colours.
• White light consists of different colours and each of these colours have different wavelength.
• Because of different wavelength of each colour it gets deviated by different angles of deviation.
• This proves each of the colours gets deviated differently.

Cauchy’s Formula

• The refractive index of the prism varies with wavelength.
• μ = (A + (B/ λ2)) where (A and B are Cauchy’s constants).
• This shows that μ ∝ (1/ λ).
• As the wavelength of violet colour is least so the refractive index would be maximum this shows it gets deviated the least.
• Also, δ∝ μ. Angle of minimum deviation will be maximum for violet, so the violet will deviate the most.

Some Natural Phenomena due to Sunlight

Rainbow

• Rainbow is a phenomenon due to combined effect of dispersion, refraction and reflection of sunlight by spherical water droplets of water.
• Rainbow appears when the sun is shining in on one part of the sky (say near western horizon) while it is raining in the opposite part of the sky (say eastern horizon).
• Sunlight is first refracted as it enters a raindrop, which causes the different wavelengths (colours) of white light to separate.
• Longer wavelength of light (red) are bent the least while the shorter wavelength (violet) are bent the most.
• The rays strike the inner surface of the water drop and get internally reflected if the angle between the refracted ray and normal to the drop surface is greater than the critical angle (48º, in this case).
• The reflected light is refracted again as it comes out of the drop as shown in the figure.
• It is found that the violet light emerges at an angle of 40º related to the incoming sunlight and red light emerges at an angle of 42º. For other colours, angles lie in between these two values.
• This lead to the formation of primary rainbow.
• The observer sees a rainbow with red colour on the top and violet on the bottom. Thus, the primary rainbow is a result of three-step process, that is, refraction, reflection and refraction.
• When light rays undergo two internal reflections inside a raindrop, instead of one as in the primary rainbow, a secondary rainbow is formed as shown in Fig. It is due to four-step process.
• The intensity of light is reduced at the second reflection and hence the secondary rainbow is fainter than the primary rainbow.

Problem:-

A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism.

The angle of minimum deviation is measured to be 40°. What is the refractive index of the material of the prism?

The refracting angle of the prism is 60°. If the prism is placed in water (refractive index 1.33), predict the new angle of minimum deviation of a parallel beam of light.

Given:

Angle of minimum deviation, δm = 40°

Angle of the prism, A = 60°

Refractive index of water, µ = 1.33

Refractive index of the material of the prism = µ’

The angle of deviation is related to refractive index (µ’) as:

µ’= (sin (A + δm)/2)/ (sin (A/2))

= (sin (600 + 400))/ (sin (600/2))

= (sin (500))/ (sin (300))

= 1.532

Hence, the refractive index of the material of the prism is 1.532.

Since the prism is placed in water,

Let δm ‘= the new angle of minimum deviation for the same prism.

The refractive index of glass with respect to water is given by the relation:

μgw = (µ’)/ (µ)

= (sin (A + δm ‘)/ (2)) / (sin (A/2))

(sin (A + δm ‘)/ (2)) = (sin (A/2)) (µ’)/ (µ)

(sin (A + δm ‘)/ (2)) = sin (600) (1.532)/ (1.33)

(sin (A + δm ‘)/ (2)) = 0.5759

(A + δm ‘)/ (2) = sin -1(0.5759)

(A + δm ‘)/ (2) = 35.160

(600 + δm ‘) = 70.320

Therefore, δm ‘= (70.320 – 600)

δm ‘=10.320

Hence, the new minimum angle of deviation is 10.32°.

Optical Instruments

• Optical instruments are instruments using reflecting and refracting properties of mirrors, lenses and prisms.
• A number of optical devices and instruments have been designed utilising reflecting and refracting properties of mirrors, lenses and prisms.
• Periscope, kaleidoscope, binoculars, telescopes; microscopes are some examples of optical devices and instruments.
• Some of optical instruments which consists of lenses and prisms are:-
• Binoculars
• Telescope
• Microscope
• Eye

Human Eye

• Our eyes are marvellous organs that have the capability to interpret incoming electromagnetic waves as images through a complex process. These are our greatest assets and we must take proper care to protect them.
• Components of Eye:-
1. Cornea
2. Aqueous Humour
3. Pupil
4. Iris
5. Lens
6. Ciliary Muscles
7. Vitreous humour
8. Retina it contains Rods and Cones.
• Light enters the eye through a curved front surface, the cornea. It passes through the pupil which is the central hole in the iris. The size of the pupil can change under control of muscles.
• The light is further focussed by the eye lens on the retina. The retina is a film of nerve fibres covering the curved back surface of the eye.
• The retina contains rods and cones which sense light intensity and colour, respectively, and transmit electrical signals via the optic nerve to the brain which finally processes this information.
• The shape (curvature) and therefore the focal length of the lens can be modified somewhat by the ciliary muscles.
• For example, when the muscle is relaxed, the focal length is about 2.5 cm and objects at infinity are in sharp focus on the retina.
• When the object is brought closer to the eye, in order to maintain the same image-lens distance (≅5 cm), the focal length of the eye lens becomes shorter by the action of the ciliary muscles.
• This property of the eye is called accommodation. If the object is too close to the eye, the lens cannot curve enough to focus the image on to the retina, and the image is blurred.
• The closest distance for which the lens can focus light on the retina is called the least distance of distinct vision, or the near point.
• The standard value for normal vision is taken as 25 cm. (Often the near point is given the symbol D.) This distance increases with age, because of the decreasing effectiveness of the ciliary muscle and the loss of flexibility of the lens.
• The near point may be as close as about 7 to 8 cm in a child ten years of age, and may increase to as much as 200 cm at 60 years of age.
• If an elderly person tries to read a book at about 25 cm from the eye, the image appears blurred. This condition (defect of the eye) is called presbyopia.
• Some Optical Defects of Eye:-
• Myopia: – The light from a distant object arriving at the eye-lens may get converged at a point in front of the retina. This type of defect is called near-sightedness or myopia.
• This means that the eye is producing too much convergence in the incident beam. To correct this, we interpose a concave lens between the eye and the object, with the diverging effect desired to get the image focussed on the retina.

Short-sighted or myopic eye and its correction

• Hypermetropia: – If the eye-lens focusses the incoming light at a point behind the retina, a convergent lens is needed to compensate for the defect in vision. This defect is called farsightedness or hypermetropia.

Farsighted or hypermetropic eye and its correction

• Astigmatism:- This occurs when the cornea is not spherical in shape.
• For example, the cornea could have a larger curvature in the vertical plane than in the horizontal plane or vice-versa. If a person with such a defect in eye-lens looks at a wire mesh or a grid of lines, focussing in either the vertical or the horizontal plane may not be as sharp as in the other plane.
• Astigmatism results in lines in one direction being well focussed while those in a perpendicular direction may appear distorted.
• Astigmatism can be corrected by using a cylindrical lens of desired radius of curvature with an appropriately directed axis. This defect can occur along with myopia or hypermetropia.

Astigmatic eye and its correction

Problem:-

A myopic person has been using spectacles of power –1.0 dioptre for distant vision.

During old age he also needs to use separate reading glass of power + 2.0 dioptres. Explain what may have happened.

Given:

Power P =-1.0D

Therefore, f = (1/P)

= – (1/1.0)

= – 1m

=-100cm

Hence, the far point of the person is 100 cm. He might have a normal near point of 25 cm. When he uses the spectacles, the objects placed at infinity produce virtual images at 100 cm. He uses the ability of accommodation of the eye-lens to see the objects placed between 100 cm and 25 cm.

During old age, the ability of accommodation is partially lost so the near point of the person recedes.

Here u = -25cm; v = -50cm;

Using, (1/f) = (1/v) – (1/u)

= (-1/50) + (1/25)

= (-1 +2)/ (50)

= (1/50)

Therefore f = 50cm

Or P = (1/f) x 100

= (1/50) x100

=+2 dioptres

Therefore, on wearing spectacles of power +2 dioptres, the image of the object lying at a distance of 25cm is formed at a distance of 50cm.

Microscope

• Microscope is an instrument that gives an enlarged image of minute object.
• There are 2 types of microscope:-
• Simple
• Compound

Simple Microscope

• An instrument that gives an enlarged image of a minute object.
• A simple magnifier or microscope is a converging lens of small focal length.
• There are 2 types of Microscopes:-
1. Simple Microscope 2. Compound Microscope

Simple Microscope

• The lens is held near the object, one focal length away or less, and the eye is positioned close to the lens on the other side.
• Image which we will get is an erect, magnified and virtual image of the object at a distance so that it can be viewed comfortably, i.e., at 25 cm or more.

To Increase Magnifying Power of Simple Microscope

• If the object is at a distance f, the image is at infinity. However, if the object is at a distance slightly less than the focal length of the lens, the image is virtual and closer than infinity.
• Although the closest comfortable distance for viewing the image is when it is at the near point (distance

D ≅ 25 cm), it causes some strain on the eye.

• Therefore, the image formed at infinity is often considered most suitable for viewing by the relaxed eye.
• Both the cases can be seen in the figures given below:
• The linear magnification m, for the image formed at the near point D, by a simple microscope can be obtained by using the relation:-
• m=(v/u) = v((1/v)-(1/f))
• =(1- (v/f))
• Using the sign conventions, v= (-) ive and same as D.
• Therefore, magnification will be m =(1 +(D/f))
• Since D is about 25 cm, to have a magnification of six, one needs a convex lens of focal length, f = 5 cm.
• Magnification when the image is at infinity.
• Suppose the object has a height h. The maximum angle it can subtend, and be clearly visible (without a lens), is when it is at the near point, i.e., a distance D.
• The angle subtended is then given by:-
• tan θ0 =(h/D)≈ θ0
• To find the angle subtended at the eye by the image when the object is at u.
• Therefore, (h’/h) = m = (v/u)
• Angle subtended by the image will be;-
• tan θ1= (h’/-v) = (h/-v) x (v/u)
• = (h/-u) ≈ θ.
• The angle subtended by the object, when it is at u=-f.
• θi =(h/f).
• The angular magnification is m =( θi/ θ0) =(D/f)

Compound Microscope

• In order to have large magnifications compound microscope is used.
• The lens nearest the object, called the objective, forms a real, inverted, magnified image of the object. This serves as the object for the second lens, the eyepiece, which functions essentially like a simple microscope or magnifier, produces the final image, which is enlarged and virtual.
• The first inverted image is thus near (at or within) the focal plane of the eyepiece, at a distance appropriate for final image formation at infinity, or a little closer for image formation at the near point.
• Clearly, the final image is inverted with respect to the original object.
• Using tanβ = (h/f0) = (h’/L)
• Magnification (mo) due to objective = (h’/h) =(L/f0)
• Where h’ = size of the first image
• h= size of the object
• fo = focal length of the objective lens
• fe= focal length of the eye-piece
• L (tube length) = Distance between focal length of the second objective lens and the first focal length of the eye-piece.
• When the final image is formed at the near point, then the angular magnification will be :-
• me=(1+(D/fe))
• When the final image is formed at infinity, the angular magnification due to the eyepiece is:-
• me = (D/fe)
• Total magnification will be given as:-
• m=(mome) =(L/f0)(D/fe)
• Note: – In order to achieve a large magnification of a small object (hence the name microscope), the objective and eyepiece should have small focal lengths. In practice, it is difficult to make the focal length much smaller than 1 cm.

Telescope

• An instrument used to view distant objects clearly.
• It consists of:- (a) Objective lens (b) Eyepiece

Working of Telescope

• The telescope is used to provide angular magnification of distant objects. The objective has a large focal length and a much larger aperture than the eyepiece because object is very far away.
• Light from a distant object enters the objective and a real and inverted image is formed at its second focal point.
• This image acts as an object for the eyepiece; it magnifies this image producing a final inverted image.

Magnification

• The magnifying power m is the ratio of the angle β subtended at the eye by the final image to the angle α which the object subtends at the lens or the eye.
• Therefore, m≈ (β / α) ≈ (h/fe) x (fo/h) = (fo/h).
• In this case, the length of the telescope tube is (fo + fe).
• In addition, a pair of inverting lenses to make the final image erect.
• Refracting telescopes can be used both for terrestrial and astronomical observations.
Chapter-9-Ray-Optics-Optical-Instruments

## Important Questions for CBSE Class 12 Physics Chapter 9 – Ray Optics and Optical Instruments

### 1 Mark Questions ( Ray Optics And Optical Instruments )

1. A person is standing before a concave mirror cannot see his image, unless he is beyond the centre of curvature? Why?

Ans. When man stands beyond focus is i.e. between focus and centre of curvature, his real and inverted image is formed beyond C is beyond him and thus he cannot see the image. But when he stands beyond C, image is formed between focus and centre of curvature is in front of him and thus he is able to see his image.

2. For what angle of incidence, the lateral shift produced by a parallel sided glass plate is maximum?

Ans. We know

D =t

Lateral shift is maximum

3. You read a newspaper, because of the light if reflects. Then why do you not see even a faint image of yourself in the newspaper?

Ans. The image is produced due to regular reflection of light but when we read a newspaper, because of diffused (irregular) reflection of light we are not able to see even a faint image.

4. A substance has critical angle of  for yellow light what is its refractive index?

Ans.

5. An object is placed between the pole and focus of a concave mirror produces a virtual and enlarged image. Justify using mirror formula?

Ans.

For a concave mirror

Given V < f so  is positive, hence image is virtual.
Now magnification

Hence enlarged image is produce

6. A converging and diverging lens of equal focal lengths are placed coaxially in contact. Find the focal length and power of the combination?

Ans.

For converging lens

For diverging lens

P=0

Hence

7. The refractive index of a material of a convex lens is n1 it is immersed in a medium of refractive index. A parallel beam of light is incident on thelens. Trace the path of the emergent rays when.

Ans. When then the convex lens behaves as a concave

8. In a telescope the focal length of the objective and the eye piece are 60cm and 5cm respectively. What is? (1) Its magnification power (2) Tube length

Ans. magnification

Tube length

9. Show the variation of u and  in case of a convex mirror?

Ans.

10. Two lenses having focal length are placed coaxially at a distance x from each other. What is the focal length of the combination?

Ans. :

11. Does short-sightedness (myopia) or long-sightedness (hypermetropia) imply necessarily that the eye has partially lost its ability of accommodation? If not, what might cause these defects of vision?

Ans. A myopic or hypermetropic person can also possess the normal ability of accommodation of the eye-lens. Myopia occurs when the eye-balls get elongated from front to back. Hypermetropia occurs when the eye-balls get shortened. When the eye-lens loses its ability of accommodation, the defect is called presbyopia.

### 2 Marks Questions ( Ray Optics And Optical Instruments )

1. What are optical fibres? Give their one use?

Ans. Optical fibres consist of thin and long strands of fine quality glass or quartz coated with a thin layer of material of refractive index less than the refractive index of strands. They work on the principle of total internal reflection so they do not suffer any loss.

Uses

The optical fibres are used in medical investigations i.e. one can examine the inside view of stomach and intestine by a method called endoscopy.

2. How the focal lengths of a lens change with increase in the wavelength of the light?

Ans.

i.e. when wavelength increases  decreases and according to
focal length increases.

3. Show with a ray diagram, how an image is produced in a total reflecting prism?

Ans. the two rays from the object PQ undergoes total internal reflection firstly at the face AB and then at BC forming the find image  (real and inverted image)

4. The radii of the curvature of the two spherical surfaces which is a lens of required focal length are not same. It forms image of an object. The surfaces of the lens facing the object and the image are inter-changed. Will the position of the image change?

Ans. As we know

When  gets interchanged focal length of the lens remains the same hence position of the image will not change.

5. A thin converging lens has focal length (f) when illuminated by violet light. State with reason how the focal length of the lens will change if violet light is replaced by red light

Ans. Since

n for violet is more that n for red colour hence focal length of the lens will decreases when violet light is replaced by red light.

6. Thin prism of angle gives a deviation of . What is the refractive index of material of the prism?

Ans.

n =1.41

7. Although the surfaces of a goggle lens are curved it does not have any power. Why?

Ans. The two surface of the goggle lens are parallel i.e. one surface convex and the other concave thus the power of the two surfaces and equal but of opposite sign.

8. A ray of light in incident normally on one face of the prism of apex angle  and refractive index . Find the angle of deviation for the ray of light?

Ans. When ray PQ falls normally on AB then if goes straight at QR (no refraction)

Applying Snell’s law for face AC

r = 45o

Now angle of deviation

9. Following data was recorded for values of object distance and corresponding values of image distance in the experiment on study of real image formation by a convex lens of power +5 D. one of three observation is incorrect. Identify and give reason?

S. No.(u)123456

Object distance253035455055

Image distance976137353230

Ans.

Observation (3) is incorrect because both object and the image here lies between f and 2f.

10. A bird flying high in the air appears to be higher than in reality. Explain why?

Ans. Bird flying in air is in the rarer medium and if we see it from denser medium than light form bird refract towards the normal thus appears to come from the higher point. i.e. Apparent height > Real height (BIRD APPEARS TO BE)

11. What is the focal length of a convex lens of focal length 30 cm in contact with a concave lens of focal length 20 cm? Is the system a converging or a diverging lens? Ignore thickness of the lenses.

Ans. Focal length of the convex lens, = 30 cm

Focal length of the concave lens, = – 20 cm

Focal length of the system of lenses = f

The equivalent focal length of a system of two lenses in contact is given as:

Hence, the focal length of the combination of lenses is 60 cm. The negative sign indicates that the system of lenses acts as a diverging lens.

12. The image of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall 3 m away by means of a large convex lens. What is the maximum possible focal length of the lens required for the purpose?

Ans. Distance between the object and the image, d= 3 m

Maximum focal length of the convex lens =

For real images, the maximum focal length is given as:

Hence, for the required purpose, the maximum possible focal length of the convex lens is 0.75 m.

13. A screen is placed 90 cm from an object. The image of the object on the screen is formed by a convex lens at two different locations separated by 20 cm. Determine the focal length of the lens.

Ans. Distance between the image (screen) and the object, D= 90 cm

Distance between two locations of the convex lens, d = 20 cm

Focal length of the lens = f

Focal length is related to and D as:

Therefore, the focal length of the convex lens is 21.39 cm.

14. You are given prisms made of crown glass and flint glass with a wide variety of angles. Suggest a combination of prisms which will

(a) deviate a pencil of white light without much dispersion,

(b) disperse (and displace) a pencil of white light without much deviation.

Ans.(a) Place the two prisms beside each other. Make sure that their bases are on the opposite sides of the incident white light, with their faces touching each other. When the white light is incident on the first prism, it will get dispersed. When this dispersed light is incident on the second prism, it will recombine and white light will emerge from the combination of the two prisms.

(b) Take the system of the two prisms as suggested in answer (a). Adjust (increase) the angle of the flint-glass-prism so that the deviations due to the combination of the prisms become equal. This combination will disperse the pencil of white light without much deviation.

15. A myopic person has been using spectacles of power -1.0 dioptre for distant vision. During old age he also needs to use separate reading glass of power + 2.0 dioptres. Explain what may have happened.

Ans. The power of the spectacles used by the myopic person, P= – 1.0 D

Focal length of the spectacles,

Hence, the far point of the person is 100 cm. He might have a normal near point of 25 cm. When he uses the spectacles, the objects placed at infinity produce virtual images at 100 cm. He uses the ability of accommodation of the eye-lens to see the objects placed between 100 cm and 25 cm.

During old age, the person uses reading glasses of power,

The ability of accommodation is lost in old age. This defect is called presbyopia. As a result, he is unable to see clearly the objects placed at 25 cm.

16. A person looking at a person wearing a shirt with a pattern comprising vertical and horizontal lines is able to see the vertical lines more distinctly than the horizontal ones. What is this defect due to? How is such a defect of vision corrected?

Ans.In the given case, the person is able to see vertical lines more distinctly than horizontal lines. This means that the refracting system (cornea and eye-lens) of the eye is not working in the same way in different planes. This defect is called astigmatism. The person’s eye has enough curvature in the vertical plane. However, the curvature in the horizontal plane is insufficient. Hence, sharp images of the vertical lines are formed on the retina, but horizontal lines appear blurred. This defect can be corrected by using cylindrical lenses.

17. A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. What is the magnifying power of the telescope for viewing distant objects when

(a) the telescope is in normal adjustment (i.e., when the final image is at infinity)?

(b) the final image is formed at the least distance of distinct vision (25 cm)?

Ans. Focal length of the objective lens,= 140 cm

Focal length of the eyepiece,  = 5 cm

Least distance of distinct vision, d = 25 cm

(a) When the telescope is in normal adjustment, its magnifying power is given as:

(b) When the final image is formed at d, the magnifying power of the telescope is given as:

18. Light incident normally on a plane mirror attached to a galvanometer coil retraces backwards as shown in Fig. 9.36. A current in the coil produces a deflection of 3.5°of the mirror. What is the displacement of the reflected spot of light on a screen placed 1.5 m away?

Ans. Angle of deflection,

Distance of the screen from the mirror, D= 1.5 m

The reflected rays get deflected by an amount twice the angle of deflection i.e.,

The displacement (d) of the reflected spot of light on the screen is given as:

Hence, the displacement of the reflected spot of light is 18.4 cm.

### 3 Marks Questions ( Ray Optics And Optical Instruments )

1. Find the radius of curvature of the convex surface of a plane convex lens, whose focal length is 0.3m and the refractive index of the material of the lens is 1.5?

Ans.

For plane convex lens

R = 0.15 m

2. Show that the limiting value of the angle of prism is twice its critical angle? Hence define critical angle?

Ans. Angle of the prism

For limiting

(Maximum)

Value of angle of prism for means

But when  = C

Amax = 2C

The angle of incidence for which angle of refraction is 90is called critical angle.

3. Draw a labeled diagram of telescope when the image is formed at the least distance of distinct vision? Hence derive the expression for its magnifying power?

Ans. magnifying power
=

(since angles are very small)

—-(i)

For eye piece

Multiply by D

Substituting in e.g. (i)

4. Drive the expression for the angle of deviation for a ray of light passing through an equilateral prism of refracting angle A?

Ans. At the surface AB

At the surface AC

—-(1)

—-(2)

Now in

Or

—-(3)

From (2) and (3)

—-(4)

Substituting equation (4) in equation (1)

= i + e – A

Or

A +  = i + e

5. Draw a ray diagram to illustrate image formation by a Newtonian type reflecting telescope? Hence state two advantages of it over refracting type telescopes?

(1) The image formed in reflecting type telescope is free from chromatic aberrations

(2) The image formed is very bright due to its large light gathering power.

NEWTONIAN TELESCOPE (REFLECTING TYPE)

6. The magnifying power of an astronomical telescope in the normal adjustment position is 100. The distance between the objective and the eye piece is 101cm. calculate the focal length of the objective and the eye piece.

Ans.  —-(1)

fo = 100 fe —-(2)

Substituting equation. (2) in equation (1)

fe + 100fe = 101

101fe = 101

fe = 1cm

Substituting fe in equation (2)

7. A convex lens made up of refractive index n1 is kept in a medium of refractive index . Parallel rays of light are incident on the lens. Complete the path of rays of light emerging from the convex lens if

(1)

(2)

(3)

Ans. (1) When the lens behaves as a convex lens.

(2) When the lens behaves as a plane plate so no refraction takes place

(3) When the lens behave as a convex lens

8. Derive the relation

Where are focal lengths of two thin lenses and F is the focal length of the combination in contact.

Ans. Consider two thin lenses in contact having focal length For the first lens

For the second lens acts as an object which forms the final image I

Using lens formula

For n no. of thin lenses is contact

9. A convex lens has a focal length 0.2m and made of glass is immersed in water  find the change in focal length of the lens?

Ans. Fair = 0.2m, ag = 1.50

—-(1)

Now

Where too is the focal length of the lens when immersed in water?

10. A reflecting type telescope has a concave reflector of radius of curvature 120cm. calculate the focal length of eye piece to achieve a magnification of 20?

Ans. M = 20

R = 120cm (fro concave reflector)

fe = 3cm

11. Show that a convex lens produces an N time magnified image, when the object distances from the lens have magnitude . Here f is the magnitude of the focal length of the lens. Hence find two values of object distance, for which value of u convex lens of power 2.5 D will produce an image that is four times as large as the object?

Ans. Magnifying power

For real image m = -N

Or

—-(1)

For virtual image m =N

Or

—-(2)

From equation (1) & (2) we can say that magnification produced by a lens can be N if u

Now power of a lens = 2.5 D

m equation (1)

u = 30cm or -50cm

12. Define total infernal reflection of light? Hence write two advantages of total reflecting prisms over a plane mirror?

Ans. The phenomenon of reflection of light when a ray of light traveling from a denser medium is sent back to the same denser medium provided the angle of incidence is greater than the angle called critical angle is called total internal reflection.

1. It does require silvering

2. Multiple reflections do not take place in a reflecting prism due to this; only one image is formed, which is very bright.

13. An equi–convex lens of radius of curvature R is cut into two equal parts by a vertical plane, so it becomes a plano-convex lens. If f is the focal length of equi–convex lens, then what will be focal length of the plano –convex lens?

Ans. We know

For equi –convex lens

—-(1)

For plano convex lens

—-(2)

From (1) & (2)

Or

f’’ = 2f

14. A converging lens of focal length 6.25cm is used as a magnifying glass if near point of the observer is 25cm from the eye and the lens is held close to the eye. Calculate (1) Distance of object from the lens. (2) Angular magnification and (3) Angular magnification when final image is formed at infinity.

Ans. (1)

u = -5cm

(2)

m = 5cm

m = 4

15. Draw a graph to show that variation of angle of deviation Dm with that of angle of incidence i for a monochromatic ray of light passing through a glass prism of refracting angle A. hence deduce the relation?

Ans. For the minimum deviation position

(Say)

We know =

Also = A

Or 2r = A

Applying minimum deviation condition is equation. (1)

2i =

Applying Snell’s law

Or

16. An object of size 3.0 cm is placed 14 cm in front of a concave lens of focal length 21 cm. Describe the image produced by the lens. What happens if the object is moved further away from the lens?

Ans. Size of the object, = 3 cm

Object distance, u= – 14 cm

Focal length of the concave lens, f = – 21 cm

Image distance = v

According to the lens formula, we have the relation:

Hence, the image is formed on the other side of the lens, 8.4 cm away from it. The negative sign shows that the image is erect and virtual.

The magnification of the image is given as:

Hence, the height of the image is 1.8 cm.

If the object is moved further away from the lens, then the virtual image will move toward the focus of the lens, but not beyond it. The size of the image will decrease with the increase in the object distance.

17. A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12 cm from P. At what point does the beam converge if the lens is (a) a convex lens of focal length 20 cm, and (b) a concave lens of focal length 16 cm?

Ans.In the given situation, the object is virtual and the image formed is real.

Object distance, u = +12 cm

(a) Focal length of the convex lens, f= 20 cm

Image distance = v

According to the lens formula, we have the relation:

Hence, the image is formed 7.5 cm away from the lens, toward its right.

(b) Focal length of the concave lens, f= –16 cm

Image distance = v

According to the lens formula, we have the relation:

Hence, the image is formed 48 cm away from the lens, toward its right.

18. Double-convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be 20cm?

Ans. Refractive index of glass,

Focal length of the double-convex lens, f = 20 cm

Radius of curvature of one face of the lens =

Radius of curvature of the other face of the lens =

Radius of curvature of the double-convex lens =R

The value of can be calculated as:

Hence, the radius of curvature of the double-convex lens is 22 cm.

19. A small telescope has an objective lens of focal length 144 cm and an eyepiece of focal length 6.0 cm. What is the magnifying power of the telescope? What is the separation between the objective and the eyepiece?

Ans. Focal length of the objective lens,  = 144 cm

Focal length of the eyepiece,  = 6.0 cm

The magnifying power of the telescope is given as:

The separation between the objective lens and the eyepiece is calculated as:

Hence, the magnifying power of the telescope is 24 and the separation between the objective lens and the eyepiece is 150 cm.

20. (a) A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eyepiece of focal length 1.0 cm is used, what is the angular magnification of the telescope?

(b) If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is , and the radius of lunar orbit is.

Ans. Focal length of the objective lens,
= 15 m = cm

Focal length of the eyepiece,
= 1.0 cm

(a) The angular magnification of a telescope is given as:

Hence, the angular magnification of the given refracting telescope is 1500.

(b) Diameter of the moon,
d

r0= m

Let be the diameter of the image of the moon formed by the objective lens.

The angle subtended by the diameter of the moon is equal to the angle subtended by the image.

Hence, the diameter of the moon’s image formed by the objective lens is 13.74 cm

21. A small pin fixed on a table top is viewed from above from a distance of 50 cm. By what distance would the pin appear to be raised if it is viewed from the same point through a 15 cm thick glass slab held parallel to the table? Refractive index of glass = 1.5. Does the answer depend on the location of the slab?

Ans. Actual depth of the pin, d= 15 cm

Apparent dept of the pin =

Refractive index of glass,

Ratio of actual depth to the apparent depth is equal to the refractive index of glass, i.e.

The distance at which the pin appears to be raised =

=15-10=5 cm

For a small angle of incidence, this distance does not depend upon the location of the slab.

22. A man with normal near point (25 cm) reads a book with small print using a magnifying glass: a thin convex lens of focal length 5 cm.

(a) What is the closest and the farthest distance at which he should keep the lens from the page so that he can read the book when viewing through the magnifying glass?

(b) What is the maximum and the minimum angular magnification (magnifying power) possible using the above simple microscope?

Ans. (a) Focal length of the magnifying glass, f = 5 cm

Least distance of distance vision, d = 25 cm

Closest object distance = u

Image distance, v= – d= – 25 cm

According to the lens formula, we have:

Hence, the closest distance at which the person can read the book is 4.167 cm.

For the object at the farthest distant (u‘), the image distance

According to the lens formula, we have:

Hence, the farthest distance at which the person can read the book is
5 cm.

(b) Maximum angular magnification is given by the relation:

Minimum angular magnification is given by the relation:

23. A card sheet divided into squares each of size 1 mm2 is being viewed at a distance of 9 cm through a magnifying glass (a converging lens of focal length 9 cm) held close to the eye.

(a) What is the magnification produced by the lens? How much is the area of each square in the virtual image?

(b) What is the angular magnification (magnifying power) of the lens?

(c) Is the magnification in (a) equal to the magnifying power in (b)? Explain.

Ans. (a) Area of each square, A= 1 mm2

Object distance, u= – 9 cm

Focal length of a converging lens, f = 10 cm

For image distance v, the lens formula can be written as:

Magnification,

∴Area of each square in the virtual image = (10)2A

= 102 ×1 = 100

= 1

(b) Magnifying power of the lens

(c) The magnification in (a) is not the same as the magnifying power in (b).

The magnification magnitude is and the magnifying power is.

The two quantities will be equal when the image is formed at the near point (25 cm).

24. (a) At what distance should the lens be held from the figure in Exercise 9.29 in order to view the squares distinctly with the maximum possible magnifying power?

(b) What is the magnification in this case?

(c) Is the magnification equal to the magnifying power in this case? Explain.

Ans. (a) The maximum possible magnification is obtained when the image is formed at the near point (d= 25 cm).

Image distance, v= – d= – 25 cm

Focal length, f= 10 cm

Object distance = u

According to the lens formula, we have:

Hence, to view the squares distinctly, the lens should be kept 7.14 cm away from them.

(b) Magnification =

(c) Magnifying power =

Since the image is formed at the near point (25 cm), the magnifying power is equal to the magnitude of magnification.

25. What should be the distance between the object in Exercise 9.30 and the magnifying glass if the virtual image of each square in the figure is to have an area of 6.25 mm2. Would you be able to see the squares distinctly with your eyes very close to the magnifier?

[Note: Exercises 9.29 to 9.31 will help you clearly understand the difference between magnification in absolute size and the angular magnification (or magnifying power) of an instrument.]

Ans. Area of the virtual image of each square,

Area of each square,

Hence, the linear magnification of the object can be calculated as:

But m =

…(1)

Focal length of the magnifying glass, f= 10 cm

According to the lens formula, we have the relation:

The virtual image is formed at a distance of 15 cm, which is less than the near point (i.e., 25 cm) of a normal eye. Hence, it cannot be seen by the eyes distinctly.

26. (a) For the telescope described in Exercise 9.34 (a), what is the separation between the objective lens and the eyepiece?

(b) If this telescope is used to view a 100 m tall tower 3 km away, what is the height of the image of the tower formed by the objective lens?

(c) What is the height of the final image of the tower if it is formed at 25 cm?

Ans. Focal length of the objective lens,  = 140 cm

Focal length of the eyepiece,  = 5 cm

(a) In normal adjustment, the separation between the objective lens and the eyepiece

(b) Height of the tower, h1= 100 m

Distance of the tower (object) from the telescope, u= 3 km = 3000 m

The angle subtended by the tower at the telescope is given as:

The angle subtended by the image produced by the objective lens is given as:

Where,

= Height of the image of the tower formed by the objective lens

Therefore, the objective lens forms a 4.7 cm tall image of the tower.

(c) Image is formed at a distance, d= 25 cm. The magnification of the eyepiece is given by the relation:

Height of the final image

Hence, the height of the final image of the tower is 28.2 cm.

### 5 Marks Questions ( Ray Optics And Optical Instruments )

1. Prove that

When refraction occurs of a convex spherical refracting surface and the ray travels

from rarer to denser medium.

Ans. From  —-(i)

Similarly form

—-(ii)

From

Same aperture of the spherical surface is small so point N lies close to P and since angles  are very small and

and

Applying sign conventional

and

Substituting these values in e.g. (1) & (2)

According to snell’s law

(Since angles are very small)

2. A lens forms a real image of an object. The distance of the object. From the lens is U cm and the distance of the image from the lens is cm. The given graph shows the variation of  and U

(a) What is the nature of the lens?

(b) Using the graph find the focal length of the lens?

(c) Draw a ray diagram to show the formation of image of same size as that of object in case of converging lens hence derive lens equation?

Ans. (a) convex lens

(b)

i.e

In the given graph f = 10cm.

(c)

and are similar

—-(2)

Combining equation (1) & (2)

Using sign conventions

Divide by Uf

Or

Hence derived

3. By stating sign conventions and assumptions used derive the relation between  in case of a concave mirror?

Ans. Sign conventions

(1) All distances are measured from the pole of the mirror.

(2) Distance measured in the direction of incident light is positive and those measured in the direction opposite to the incident light are negative.

(3) Height measured upwards is positive and height measured downwards is negative.

Assumptions

(1) Aperture of the spherical mirror is considered to be very small.

Remaining angles equal

—-(1) (Dlies closetol AB + MD)

Similarly  and are also similar

—-(2)

Combining equation (1) & (2) and using sign conventions

Divide by we  get

4. (a) A person looking at a mesh of crossed wires is able to see the vertical lines more distinctly than the horizontal wires. What is the effect due to? How is such a defect of vision corrected?

(b) A man with normal near point (25cm) reads a book with small print using a magnifying glass: a thin convex lens of focal length 5cm.

(i) What is the closest and the farthest distance at which he can read the book when viewing through the magnifying glass ?

(ii) What is the maximum and minimum angular magnificent (magnifying power) possible using the above simple microscope?

Ans. (a) It is due to the defect called astigmatisms and is caused due to irregular surface of cornea and curvature of the eye lens is different in different planes. This type of defect can be corrected using cylindrical lens.

(b) (i) Here f = 5cm

For closest point

and thus

u = -4.2cm

A for farthest point f = 5cm

u = -5cm

(b) (ii) Angular magnification

Maximum angular magnification

Minimum angular magnification

5. Four double convex lens with following specification are available.

(a) Which of the given four lenses should be selected as objective and eyepiece to construct an astronomical telescope and why? What will be the magnifying power and length of the tube of this telescope?

(b) An object is seen with the help of a simple microscope, firstly in red light and then is blue light. Will the magnification be same in both the cases? Why?

Ans. (a) objective of the telescope should be of large aperture as it has to gather maximum light and should be of large focal length to have maximum magnification.

Hence lens A is selected as objective and lens D as eyepiece of small aperture and small focal length.

M.P. = 20

(b) L = fo + fe

L = 100 + 5

L = 10.5 cm

When red light as replaced by blue light, magnifying power increases.

6. A small candle, 2.5 cm in size is placed at 27 cm in front of a concave mirror of radius of curvature 36 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved?

Ans. Size of the candle, h= 2.5 cm

Image size = h

Object distance, u= – 27 cm

Radius of curvature of the concave mirror, R= – 36 cm

Focal length of the concave mirror,

Image distance = v

The image distance can be obtained using the mirror formula:

Therefore, the screen should be placed 54 cm away from the mirror to obtain a sharp image.

The magnification of the image is given as:

The height of the candle’s image is 5 cm. The negative sign indicates that the image is inverted and real.

If the candle is moved closer to the mirror, then the screen will have to be moved away from the mirror in order to obtain the image.

7. A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror.

Ans. Height of the needle, = 4.5 cm

Object distance, u = – 12 cm

Focal length of the convex mirror, f= 15 cm

Image distance = v

The value of v can be obtained using the mirror formula:

Hence, the image of the needle is 6.7 cm away from the mirror. Also, it is on the other side of the mirror. The image size is given by the magnification formula:

Hence, magnification of the image,

The height of the image is 2.5 cm. The positive sign indicates that the image is erect, virtual, and diminished.

If the needle is moved farther from the mirror, the image will also move away from the mirror, and the size of the image will reduce gradually.

8. A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again?

Ans. Actual depth of the needle in water, = 12.5 cm

Apparent depth of the needle in water, = 9.4 cm

Refractive index of water =

The value of  can be obtained as follows:

Hence, the refractive index of water is about 1.33.

Water is replaced by a liquid of refractive index,

The actual depth of the needle remains the same, but its apparent depth changes. Let be the new apparent depth of the needle. Hence, we can write the relation:

Hence, the new apparent depth of the needle is 7.67 cm. It is less than h2. Therefore, to focus the needle again, the microscope should be moved up.

∴Distance by which the microscope should be moved up = 9.4 – 7.67

= 1.73 cm

9. Figures 9.34 (a) and (b) show refraction of a ray in air incident at  with the normal to a glass-air and water-air interface, respectively. Predict the angle of refraction in glass when the angle of incidence in water is  with the normal to a water-glass interface [Fig. 9.34(c)].

Ans. As per the given figure, for the glass – air interface:

Angle of incidence, i =

Angle of refraction, r

The relative refractive index of glass with respect to air is given by Snell’s law as:

…(1)

…(i)

As per the given figure, for the air – water interface:

Angle of incidence, i

Angle of refraction, r

The relative refractive index of water with respect to air is given by Snell’s law as:

…(2)

Using (1) and (2), the relative refractive index of glass with respect to water can be obtained as:

The following figure shows the situation involving the glass – water interface.

Angle of incidence, i

Angle of refraction = r

From Snell’s law, rcan be calculated as:

Hence, the angle of refraction at the water – glass interface is

10. A small bulb is placed at the bottom of a tank containing water to a depth of 80 cm. What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is 1.33. (Consider the bulb to be a point source.)

Ans.

Actual depth of the bulb in water, = 80 cm = 0.8 m

Refractive index of water,

The given situation is shown in the following figure:

Where,

i= Angle of incidence

r= Angle of refraction =

Since the bulb is a point source, the emergent light can be considered as a circle of radius,

Using Snell’ law, we can write the relation for the refractive index of water as:

Using the given figure, we have the relation:

∴ Area of the surface of water

Hence, the area of the surface of water through which the light from the bulb can emerge is approximately 2.61.

11. A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. The angle of minimum deviation is measured to be. What is the refractive index of the material of the prism? The refracting angle of the prism is. If the prism is placed in water (refractive index 1.33), predict the new angle of minimum deviation of a parallel beam of light.

Ans.Angle of minimum deviation, =

Angle of the prism, A =

Refractive index of water, µ= 1.33

Refractive index of the material of the prism =

The angle of deviation is related to refractive indexas:

Hence, the refractive index of the material of the prism is 1.532.

Since the prism is placed in water, let be the new angle of minimum deviation for the same prism.

The refractive index of glass with respect to water is given by the relation:

Hence, the new minimum angle of deviation is.

12. A compound microscope consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm. How far from the objective should an object be placed in order to obtain the final image at (a) the least distance of distinct vision (25 cm), and (b) at infinity? What is the magnifying power of the microscope in each case?

Ans. Focal length of the objective lens, = 2.0 cm

Focal length of the eyepiece, = 6.25 cm

Distance between the objective lens and the eyepiece, d= 15 cm

(a) Least distance of distinct vision,

∴Image distance for the eyepiece, = – 25 cm

Object distance for the eyepiece =

According to the lens formula, we have the relation:

Image distance for the objective lens,

Object distance for the objective lens =

According to the lens formula, we have the relation:

Magnitude of the object distance, = 2.5 cm

The magnifying power of a compound microscope is given by the relation:

Hence, the magnifying power of the microscope is 20.

(b) The final image is formed at infinity.

∴Image distance for the eyepiece,

Object distance for the eyepiece =

According to the lens formula, we have the relation:

Image distance for the objective lens,

Object distance for the objective lens =

According to the lens formula, we have the relation:

Magnitude of the object distance, = 2.59 cm

The magnifying power of a compound microscope is given by the relation:

Hence, the magnifying power of the microscope is 13.51.

13. A person with a normal near point (25 cm) using a compound microscope with objective of focal length 8.0 mm and an eyepiece of focal length 2.5 cm can bring an object placed at 9.0 mm from the objective in sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope,

Ans. Focal length of the objective lens,  = 8 mm = 0.8 cm

Focal length of the eyepiece,  = 2.5 cm

Object distance for the objective lens,  = – 9.0 mm = – 0.9 cm

Least distance of distant vision, d = 25 cm

Image distance for the eyepiece,  = – d= – 25 cm

Object distance for the eyepiece =

Using the lens formula, we can obtain the value of as:

We can also obtain the value of the image distance for the objective lens using the lens formula.

The distance between the objective lens and the eyepiece

The magnifying power of the microscope is calculated as:

Hence, the magnifying power of the microscope is 88.

14. Use the mirror equation to deduce that: ( Ray Optics And Optical Instruments )

(a) an object placed between f and 2f of a concave mirror produces a real image beyond 2f.

(b) a convex mirror always produces a virtual image independent of the location of the object.

(c) the virtual image produced by a convex mirror is always diminished in size and is located between the focus and the pole.

(d) an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image.

[Note: This exercise helps you deduce algebraically properties of

images that one obtains from explicit ray diagrams.]

Ans. (a) For a concave mirror, the focal length (f)is negative.

f < 0

When the object is placed on the left side of the mirror, the object distance (u)is negative.

u< 0

For image distance v, we can write the lens formula as:

…(i)

The object lies between and 2f.

(u and f are negative)

…(2)

Using equation (1), we get:

∴ is negative, i.e., vis negative.

Therefore, the image lies beyond 2f.

(b) For a convex mirror, the focal length (f) is positive.

∴ f > 0

When the object is placed on the left side of the mirror, the object distance (u) is negative.

∴ u < 0

For image distance v, we have the mirror formula:

Using equation (2), we can conclude that:

Thus, the image is formed on the back side of the mirror.

Hence, a convex mirror always produces a virtual image, regardless of the object distance.

(c) For a convex mirror, the focal length (f) is positive.

When the object is placed on the left side of the mirror, the object distance (u) is negative,

For image distance v, we have the mirror formula:

But we have u <0

Hence, the image formed is diminished and is located between the focus (f) and the pole.

(d) For a concave mirror, the focal length (f) is negative.

When the object is placed on the left side of the mirror, the object distance (u) is negative.

It is placed between the focus (f) and the pole.

For image distance v, we have the mirror formula:

The image is formed on the right side of the mirror. Hence, it is a virtual image.

For u < 0 and v > 0, we can write:

15. (a) Figure 9.35 shows a cross-section of a ‘light pipe’ made of a glass fibre of refractive index 1.68.

The outer covering of the pipe is made of a material of refractive index 1.44. What is the range of the angles of the incident rays with the axis of the pipe for which total reflections inside the pipe take place, as shown in the figure.

(b) What is the answer if there is no outer covering of the pipe?

Ans.(a) Refractive index of the glass fibre,

Refractive index of the outer covering of the pipe, = 1.44

Angle of incidence = i

Angle of refraction = r

Angle of incidence at the interface = i

The refractive index  of the inner core – outer core interface is given as:

For the critical angle, total internal reflection (TIR) takes place only when, i.e.,

Maximum angle of reflection,

Let, be the maximum angle of incidence.

The refractive index at the air – glass interface,

We have the relation for the maximum angles of incidence and reflection as:

Thus, all the rays incident at angles lying in the range will suffer total internal reflection.

(b) If the outer covering of the pipe is not present, then:

Refractive index of the outer pipe,= Refractive index of air =1

For the angle of incidence, we can write Snell’s law at the air – pipe interface as:

Since > r, All incident rays will suffer total internal reflection

(a) You have learnt that plane and convex mirrors produce virtual images of objects. Can they produce real images under some circumstances? Explain.

(b) A virtual image, we always say, cannot be caught on a screen.

Yet when we ‘see’ a virtual image, we are obviously bringing it on to the ‘screen’ (i.e., the retina) of our eye. Is there a contradiction?

(c) A diver under water, looks obliquely at a fisherman standing on the bank of a lake. Would the fisherman look taller or shorter to the diver than what he actually is?

(d) Does the apparent depth of a tank of water change if viewed obliquely? If so, does the apparent depth increase or decrease?

(e) The refractive index of diamond is much greater than that of ordinary glass. Is this fact of some use to a diamond cutter?

Ans.(a) Yes

Plane and convex mirrors can produce real images as well. If the object is virtual, i.e., if the light rays converging at a point behind a plane mirror (or a convex mirror) are reflected to a point on a screen placed in front of the mirror, then a real image will be formed.

(b) No

A virtual image is formed when light rays diverge. The convex lens of the eye causes these divergent rays to converge at the retina. In this case, the virtual image serves as an object for the lens to produce a real image.

(c) The diver is in the water and the fisherman is on land (i.e., in air). Water is a denser medium than air. It is given that the diver is viewing the fisherman. This indicates that the light rays are travelling from a denser medium to a rarer medium. Hence, the refracted rays will move away from the normal. As a result, the fisherman will appear to be taller.

(d) Yes; Decrease

The apparent depth of a tank of water changes when viewed obliquely. This is because light bends on travelling from one medium to another. The apparent depth of the tank when viewed obliquely is less than the near-normal viewing.

(e) Yes

The refractive index of diamond (2.42) is more than that of ordinary glass (1.5). The critical angle for diamond is less than that for glass. A diamond cutter uses a large angle of incidence to ensure that the light entering the diamond is totally reflected from its faces. This is the reason for the sparkling effect of a diamond.

17.(a) Determine the ‘effective focal length’ of the combination of the two lenses in Exercise 9.10, if they are placed 8.0 cm apart with their principal axes coincident. Does the answer depend on which side of the combination a beam of parallel light is incident? Is the notion of effective focal length of this system useful at all?

(b) An object 1.5 cm in size is placed on the side of the convex lens in the arrangement (a) above. The distance between the object and the convex lens is 40 cm. Determine the magnification produced by the two-lens system, and the size of the image.

Ans.Focal length of the convex lens, = 30 cm

Focal length of the concave lens, = – 20 cm

Distance between the two lenses, d= 8.0 cm

(a) When the parallel beam of light is incident on the convex lens first:

According to the lens formula, we have:

Where,

= Object distance = ∞

v1= Image distance

The image will act as a virtual object for the concave lens.

Applying lens formula to the concave lens, we have:

Where,

= Object distance

= (30 – d) = 30 – 8 = 22 cm

= Image distance

The parallel incident beam appears to diverge from a point that is from the centre of the combination of the two lenses.

(ii) When the parallel beam of light is incident, from the left, on the concave lens first:

According to the lens formula, we have:

Where,

= Object distance =

= Image distance

The image will act as a real object for the convex lens.

Applying lens formula to the convex lens, we have:

Where,

= Object distance

= – (20 + d) = – (20 + 8) = – 28 cm

= Image distance

Hence, the parallel incident beam appear to diverge

18.At what angle should a ray of light be incident on the face of a prism of refracting angle so that it just suffers total internal reflection at the other face? The refractive index of the material of the prism is 1.524.

Ans.The incident, refracted, and emergent rays associated with a glass prism ABC are shown in the given figure.

Angle of prism,

Refractive index of the prism, µ= 1.524

= Incident angle

= Refracted angle

= Angle of incidence at the face AC

e= Emergent angle =

According to Snell’s law, for face AC, we can have:

It is clear from the figure that angle

According to Snell’s law, we have the relation:

Hence, the angle of incidence is

19.For a normal eye, the far point is at infinity and the near point of distinct vision is about 25cm in front of the eye. The cornea of the eye provides a converging power of about 40 dioptres, and the least converging power of the eye-lens behind the cornea is about 20 dioptres. From this rough data estimate the range of accommodation (i.e., the range of converging power of the eye-lens) of a normal eye.

Ans.Least distance of distinct vision, d= 25 cm

Far point of a normal eye,

Converging power of the cornea,

Least converging power of the eye-lens,

To see the objects at infinity, the eye uses its least converging power.

Power of the eye-lens,  = 40 + 20 = 60 D

Power of the eye-lens is given as:

To focus an object at the near point, object distance (u)= – d= – 25 cm

Focal length of the eye-lens = Distance between the cornea and the retina

= Image distance

Hence, image distance,

According to the lens formula, we can write:

Where,

= Focal length

∴Power of the eye-lens = 64 – 40 = 24 D

Hence, the range of accommodation of the eye-lens is from 20 D to 24 D.

(a) The angle subtended at the eye by an object is equal to the angle subtended at the eye by the virtual image produced by a magnifying glass. In what sense then does a magnifying glass provide angular magnification?

(b) In viewing through a magnifying glass, one usually positions one’s eyes very close to the lens. Does angular magnification change if the eye is moved back?

(c) Magnifying power of a simple microscope is inversely proportional to the focal length of the lens. What then stops us from using a convex lens of smaller and smaller focal length and achieving greater and greater magnifying power?

(d) Why must both the objective and the eyepiece of a compound microscope have short focal lengths?

(e) When viewing through a compound microscope, our eyes should be positioned not on the eyepiece but a short distance away from it for best viewing. Why? How much should be that short distance between the eye and eyepiece?

Ans.(a)Though the image size is bigger than the object, the angular size of the image is equal to the angular size of the object. A magnifying glass helps one see the objects placed closer than the least distance of distinct vision (i.e., 25 cm). A closer object causes a larger angular size. A magnifying glass provides angular magnification. Without magnification, the object cannot be placed closer to the eye. With magnification, the object can be placed much closer to the eye.

(b) Yes, the angular magnification changes. When the distance between the eye and a magnifying glass is increased, the angular magnification decreases a little. This is because the angle subtended at the eye is slightly less than the angle subtended at the lens. Image distance does not have any effect on angular magnification.

(c) The focal length of a convex lens cannot be decreased by a greater amount. This is because making lenses having very small focal lengths is not easy. Spherical and chromatic aberrations are produced by a convex lens having a very small focal length.

(d) The angular magnification produced by the eyepiece of a compound microscope is

Where,

= Focal length of the eyepiece

It can be inferred that if  is small, then angular magnification of the eyepiece will be large.

The angular magnification of the objective lens of a compound microscope is given as

Where,

= Object distance for the objective lens

= Focal length of the objective

The magnification is large when >. In the case of a microscope, the object is kept close to the objective lens. Hence, the object distance is very little. Since is small, will be even smaller. Therefore, and are both small in the given condition.

(e)When we place our eyes too close to the eyepiece of a compound microscope, we are unable to collect much refracted light. As a result, the field of view decreases substantially. Hence, the clarity of the image gets blurred.

The best position of the eye for viewing through a compound microscope is at the eye-ring attached to the eyepiece. The precise location of the eye depends on the separation between the objective lens and the eyepiece.

21.An angular magnification (magnifying power) of 30X is desired using an objective of focal length 1.25 cm and an eyepiece of focal length 5 cm. How will you set up the compound microscope?

Ans.Focal length of the objective lens,= 1.25 cm

Focal length of the eyepiece,  = 5 cm

Least distance of distinct vision, d= 25 cm

Angular magnification of the compound microscope = 30X

Total magnifying power of the compound microscope, m= 30

The angular magnification of the eyepiece is given by the relation:

The angular magnification of the objective lens (mo) is related to meas:

m

We also have the relation:

Applying the lens formula for the objective lens:

The object should be placed 1.5 cm away from the objective lens to obtain the desired magnification.

Applying the lens formula for the eyepiece:

Where,

= Image distance for the eyepiece = – d= – 25 cm

= Object distance for the eyepiece

Separation between the objective lens and the eyepiece

=4.17+7.5

=11.67 cm

Therefore, the separation between the objective lens and the eyepiece should be 11.67 cm.

22.An angular magnification (magnifying power) of 30X is desired using an objective of focal length 1.25 cm and an eyepiece of focal length 5 cm. How will you set up the compound microscope?

Ans.Focal length of the objective lens,= 1.25 cm

Focal length of the eyepiece,  = 5 cm

Least distance of distinct vision, d= 25 cm

Angular magnification of the compound microscope = 30X

Total magnifying power of the compound microscope, m= 30

The angular magnification of the eyepiece is given by the relation:

The angular magnification of the objective lens (mo) is related to meas:

mm

Applying the lens formula for the objective lens:

And

The object should be placed 1.5 cm away from the objective lens to obtain the desired magnification.

Applying the lens formula for the eyepiece:

Where,

= Image distance for the eyepiece = – d= – 25 cm

= Object distance for the eyepiece

Separation between the objective lens and the eyepiece

=4.17+7.5

=11.67 cm

Therefore, the separation between the objective lens and the eyepiece should be 11.67 cm.

23.A Cassegrain telescope uses two mirrors as shown in Fig. 9.33. Such a telescope is built with the mirrors 20 mm apart. If the radius of curvature of the large mirror is 220 mm and the small mirror is 140 mm, where will the final image of an object at infinity be?

Ans.The following figure shows a Cassegrain telescope consisting of a concave mirror and a convex mirror.

Distance between the objective mirror and the secondary mirror, d= 20 mm

Radius of curvature of the objective mirror,

Hence, focal length of the objective mirror,

Radius of curvature of the secondary mirror, R1 = 140 mm

Hence, focal length of the secondary mirror,

The image of an object placed at infinity, formed by the objective mirror, will act as a virtual object for the secondary mirror.

Hence, the virtual object distance for the secondary mirror,

=110-20

=90 mm

Applying the mirror formula for the secondary mirror, we can calculate image distance (v)as:

Hence, the final image will be formed 315 mm away from the secondary mirror.

24.Figure 9.37 shows an equiconvex lens (of refractive index 1.50) in contact with a liquid layer on top of a plane mirror. A small needle with its tip on the principal axis is moved along the axis until its inverted image is found at the position of the needle. The distance of the needle from the lens is measured to be 45.0 cm. The liquid is removed and the experiment is repeated. The new distance is measured to be 30.0 cm. What is the refractive index of the liquid?

Ans.Focal length of the convex lens, = 30 cm

The liquid acts as a mirror. Focal length of the liquid =

Focal length of the system (convex lens + liquid), f = 45 cm

For a pair of optical systems placed in contact, the equivalent focal length is given as:

Let the refractive index of the lens be and the radius of curvature of one surface be R. Hence, the radius of curvature of the other surface is – R.

R can be obtained using the relation:

Let be the refractive index of the liquid.

Radius of curvature of the liquid on the side of the plane mirror =

Radius of curvature of the liquid on the side of the lens, R= – 30 cm

The value of can be calculated using the relation:

Ray Optics And Optical Instruments