## Rajasthan Board RBSE Class 12 Physics Chapter 11 Ray Optics

**RBSE Class 12 Physics Chapter 11 Text Book Exercise with Answers**

### RBSE Class 12 Physics Chapter 11 Multiple Choice Type Questions

Question 1.

In spherical mirrors, we consider only paraxial rays for forming the images, because :

(a) They are easy to work for geometrical purpose

(b) They consist of mostly intense incident light

(c) They form point image of the point object

(d) They produce minimum dispersion

Answer:

(c) They form point image of the point object

For paraxial rays deviation is least so we consider only paraxial rays for forming the images.

Question 2.

An object is placed at 30 cm distance from a concave mirror of focal length 20 cm, then nature, magnification of the image will be :

(a) Real, -2

(b) Apparent, -2

(c) Real, +2

(d) Apparent, +2

Answer:

(a) Real, -2

or m = -2

Question 3.

The refractive index for infrared rays :

(a) is equal to ultraviolet rays

(b) is equal to red colour rays

(c) less than ultraviolet rays

(d) greater than ultraviolet rays

Answer:

(b) is equal to red colour rays

The refractive index for infrared rays is equal to red colour rays.

Question 4.

Total internal reflection occurs, when :

(a) Light travels from optically rarer medium to optically denser medium

(b) Light travels from optically denser medium to optically rarer medium

(c) Refractive indices of both medium are equal

(d) Refractive indices of both medium are different

Answer:

(b) Light travels from optically denser medium to optically rarer medium

Total internal reflection occurs, when light travels from optically denser medium to optically rarer medium.

Question 5.

When an object is placed at a distance of 20 cm from concave lens, then small image is formed. Then the correct statement will be :

(a) Image will be inverted

(b) Image may be real

(c) Image is formed at 20 cm distance

(d) The focal length of the lens may be less than 20 cm

Answer:

(d) The focal length of the lens may be less than 20 cm

The focal length of the lens may be less than 20 cm.

Question 6.

A convex lens of power +6 D is in contact with a concave lens of power -4 D. What will be the focus length and nature of the combined lens?

(a) Concave, 25 cm

(b) Convex, 50 cm

(c) Concave, 20 cm

(d) Convex, 100 cm

Answer:

(b) Convex, 50 cm

P_{1} = +6D, P_{2} = -4 D, P = ?, f = ?

P_{eff} = P_{1} + P_{2} = 6 – 4 = +2D

∴ Focal length of combined lens

f = \(\frac{100}{P}=\frac{100}{+2}\) = +50 cm

Question 7.

A ray passes through on equilateral triangle in such a way that its angle of incidence and angle of emergence are equal and this angle is 3/4 of angle of prism. Then angle of deviation will be :

(a) 45°

(b) 70°

(c) 39°

(d) 30°

Answer:

(d) 30°

For equilateral triangle A = 60°

∴ Angle of incidence = emergent angle

So, this is condition of minimum deviation.

Question 8.

The image formed by objective lens of compound telescope will be :

(a) Virtual and bigger

(b) Virtual and small

(c) Real and point size

(d) Real and bigger

Answer:

(d) Real and bigger

The image formed by objective lens of compound telescope will be real and bigger.

Question 9.

A covexo-convex lens of refractive index 1.47 is dipped in a liquid, then it behaves like a simple plane sheet of glass. It means, the refractive index of liquid is :

(a) Greater than the refractive index of glass

(b) Less than the refractive index of glass

(c) Equal to the refractive index of glass

(d) Less than one

Answer:

(c) Equal to the refractive index of glass

The refractive index of liquid is equal to the refractive index of glass.

Question 10.

The angle of minimum deviation of a prism will be equal to its angle of refractive index. If refractive index of prism is :

(a) Between \(\sqrt{2}\) and 2

(b) Less than 1

(c) Greater than 2

(d) Between \(\sqrt{2}\) and 1

Answer:

(a) Between \(\sqrt{2}\) and 2

Question 11.

A ray of light falling normally on a plane mirror, then angle of reflection will be :

(a) 90°

(b) 180°

(c) 0°

(d) 45°

Answer:

(c) 0°

For normal incidence on a plane mirror, the angle of reflection will be zero.

Question 12.

The focal length of a concave mirror is 20 cm. An object is placed at distance 20 cm from mirror. Its image will be formed :

(a) At 2f

(b) At f

(c) At O

(d) At ∞

Answer:

(d) At ∞

When object is placed at focal length, then image will be formed at infinity.

Question 13.

An observer is watching the Stars to be twinkle on Earth. The cause is :

(a) It is true that Stars do not emit light continuously

(b) Frequency absorption of light of Stars by atmosphere of their own

(c) Frequency absorption of light of Stars by atmosphere of Earth

(d) Increase and decrease of the refractive index in atmosphere of Earth

Answer:

(d) Increase and decrease of the refractive index in atmosphere of Earth

Twinkling of Stars is possible due to increase or decrease of refractive index in atmosphere of Earth.

Question 14.

If yellow light is refracted at angle of minimum deviation from prism, then :

(a) Angle of incidence and angle of emergence are equal

(b) The sum of the angle of incidence and angle of emergence is 90°

(c) Angle of incidence is less than angle of emergence

(d) Angle of incidence is greater than angle of emergence

Answer:

(a) Angle of incidence and angle of emergence are equal

At minimum deviation, the angle of incidence and angle of emergence are equal.

Question 15.

For a healthy eye, the least distance of distinct vision and maximum distance will be :

(a) 25 cm and 100 cm

(b) 25 cm and infinite

(c) 100 cm and infinite

(d) Zero and zero to infinite

Answer:

(b) 25 cm and infinite

For a healthy eye, the least distance of distinct vision is 25 cm and maximum distance is infinite.

Question 16.

The length of a simple astronomical telescope is equal to :

(a) Difference between focal length of two lenses

(b) Half of the sum of focal distances

(c) Sum of the focal distances

(d) Multiplication of the focal distances

Answer:

(c) Sum of the focal distances

Length of a simple astronomical telescope

L = f_{0} + f_{e}

Question 17.

Apparent image of size greater than object may be formed by :

(a) Convex mirror

(b) Concave mirror

(c) Plane mirror

(d) Concave lens

Answer:

(b) Concave mirror

Concave mirror forms apparent image of size greater than object.

Question 18.

In compound microscope the final image is formed :

(a) Real and erect

(b) Virtual and inverted

(c) Virtual and erect

(d) Real and inverted

Answer:

(b) Virtual and inverted

In compound microscope the final image is virtual and inverted.

Question 19.

In reflecting type telescope, the objective used is :

(a) Convex lens

(b) Convex mirror

(c) Prism

(d) Concave mirror

Answer:

(d) Concave mirror

In reflecting type telescope, the object used is concave mirror.

Question 20.

The power of the objective lens and eyepiece of an astronomical telescope are 5 D and 20 D. They form image at infinity. What will be the magnifying power of telescope?

(a) 4

(b) 2

(c) 100

(d) 0.25

Answer:

(a) 4

Question 21.

The power of convex lens is :

(a) Negative

(b) Positive

(c) Zero

(d) Imaginary

Answer:

(b) Positive

The power of convex lens is positive.

### RBSE Class 12 Physics Chapter 11 Very Short Answer Type Questions

Question 1.

What is the focal length of a plane mirror?

Answer:

Infinite.

Question 2.

Which lens has magnification less than 1?

Answer:

In concave lens, image formed is always erect and smaller than object, so magnification of lens is always

Question 3.

What is the cause of refraction?

Answer:

The main cause of refraction is that speed of light is different in different medium.

Question 4.

What is the cause of mirage in desert area?

Answer:

Total internal reflection.

Question 5.

For equal angle of incidence, the angle of refraction for three medium A B and C are 15°, 25° and 35°. In which medium the velocity of light will be minimum?

Answer:

Refractive index of medium

µ = \(\frac{\sin i}{\sin r}\)

Here, sin i = constant

∴ µ ∝ \(\frac{1}{\sin r}\)

∴ r_{A} = 15°, r_{B} = 25°, r_{C} = 35°

∴ sinr_{A} < sinr_{B} < sinr_{C}

∴ µ_{A} > µ_{B} > µ_{C}

So, velocity of light is minimum in medium A

Question 6.

Name the principle on which optical fibre works.

Answer:

Total internal reflection.

Question 7.

At the position of minimum deviation what is relation between angle of incidence and angle of emergence?

Answer:

In the condition of minimum deviation angle of incidence = angle of emergence.

Question 8.

A convex lens is in contact with a concave lens. Focal length of both lenses are equal. What is the focal length of the combination?

Answer:

Question 9.

Sun looks reddish at sunset or sunrise. Why?

Answer:

Due to scattering of light at sunset and sunrise the sun appears to be red.

Question 10.

What is the cause of rainbow?

Answer:

The reason of formation of rainbow is the dispersion of sunlight by small water drops present in the atmosphere.

Question 11.

What is myopia? How can it be removed?

Answer:

It is a defect of vision in which nearby object are seen clearly but far object are not seen clearly. A myopic eye is corrected by using a concave lens of focal length equal to the distance of the far point F from the eye.

Question 12.

Name the factor on which the scattering of light depends.

Answer:

Scattering of light depends on wavelength.

Question 13.

What type of lens is used in simple microscope?

Answer:

In simple microscope a convex lens of less focal length is used.

Question 14.

How can you distinguish between compound microscope and telescope by observation only?

Answer:

In compound microscope, the aperture of objective lens is smaller than eyepiece, while in telescope the aperture of objective is greater than aperture of eyepiece.

### RBSE Class 12 Physics Chapter 11 Short Answer Type Questions

Question 1.

An object AB is placed in front of a concave mirror as shown in figure.

(i) Complete the ray diagram for the formation of image. 5

(ii) If the lower half of the reflecting surface of the mirror is painted black, then what will be effect on position and intensity of the image?

Answer:

(i) The ray diagram is shown in figure.

(ii) If the lower half of the reflecting surface of the mirror is painted black, then there is no effect on image but the intensity of the image will be reduce to half.

Question 2.

Write uses of spherical mirros.

Answer:

There are two types of spherical

(a) Concave mirrors

(b) Convex mirrors

(a) Uses of Concave Mirrors :

- A concave mirror is used as shaving or make up mirror because it forms a magnified and erect image of the face when it is held closer to the face.
- Doctors use concave mirrors as head mirror. The mirror is strapped to the doctor’s forhead and light from a lamp after reflection from the mirror is focussed into the throat or ear of the patient.
- A small concave mirror with a small hole at its centre is used in the doctor’s Ophthalmoscope. The doctor looks through the hole from behind the mirror while a beam of light from a lamp reflected from it is directed into the pupil of patient’s eye which makes the retina visible.
- Concave mirrors are used as reflectors in headlights of cars, etc. The source is placed at the focus of a concave mirror. The light rays after reflection travel over a large distance as a parallel intense beam.

(b) Uses of Convex Mirrors :

A convex mirror is used as a rear view mirror in automobiles. The reason is that it always forms a small and erect image and it has a larger field of view than that of a plane mirror of the same size.

Question 3.

What is relation between focal length and radius of curvature for a mirror? Derive the relation.

Answer:

Relation between focal length and radius of curvature of a spherical mirror

(i) For Convex Mirror : Suppose the focal length of a convex mirror is f and its radius of curvature is R. OA is an incident ray parallel to principal axis and AS is reflected ray which seems to come from focus F. AN is a normal to principal axis. From the law of reflection :

i = r = θ (let)

In ∆AAP; ∠FAC = ∠LAS = 0

(because both are vertically opposite angles)

and ∠ACF = ∠OAL = θ (because both are corresponding angles)

In ∆AFC,

Exterior angle AFP = Interior angle (∠FAC+∠FCA)

∠AFP = θ + θ = 2θ

From right angled ∆ANC,

tanθ = \(\frac{A N}{N C}\)

If θ is very small, then

(i) tanθ = 0 and (ii) point N is very close to P. So NC ≈ PC.

Question 4.

(i) Sun looks reddish at sunset and sunrise. Why? Explain.

(ii) For which colour the refractive index of the prism is maximum and minimum?

Answer:

(i) At the time of Sunrise and Sunset the Sun appears to be reddish : According to Rayleigh’s scattering law, the intensity of scattered light is,

I_{S} ∝ \(\frac{1}{\lambda^{4}}\)

Clearly, smaller the wavelength more is the scattering and vice-versa.

When the Sun is near the horizon at sun-set or sun-rise, the light rays have to travell a larger thickness of the atmosphere than when the Sun is overhead at noon. In accordance with Rayleigh’s scattering law, the lower wavelengths in the blue region are almost completely scattered away by the air molecules. The higher wavelengths in the red region are least scattered and reach our eyes. Hence the Sun appears almost reddish at sunset and sunrise.

(ii) Refractive index of prism is maximum for violet colour and minimum for red colour.

∴ µ_{V} = µ_{R}

Question 5.

(i) What is relation between critical angle and refractive index for a substance?

(ii) Does critical angle depend on colour of light?

Answer:

(i) Refractive index of denser medium w.r.t. rarer medium

_{r}µ_{d} = \(\frac{1}{\sin i_{c}}\)

where, i_{c} = critical angle

(ii) sin i_{c} = \(\frac{1}{\mu}\)

Refractive index µ depends on colour of light, so critical angle also depends on colour of light.

Question 6.

On what factors the focal length of the lens depend?

Answer:

\(\frac{1}{f}=(n-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)

So, f depends on following factors :

- Radius of curvature of lens.
- Refractive index of lens material.
- Refractive index of medium in which lens is placed.

Question 7.

How can you increase magnifying power of compound microscope?

Answer:

For compound microscope, f0 and fe should be small. To enlarge visible region f_{0} < f_{e}.

Question 8.

What do you mean by scattering of light? Write its use in daily life.

Answer:

Scattering of Light : When the light is incident on a molecule or very small particle the size of molecule is very small in comparison to wavelength (k) of light, then the light is absorbed by the particle and later on it is re-emitted in all possible directions. This phenomenon of re-radiation of light is known as the ‘scattering’. Thus, “Scattering of light is the phenomenon is which light is deviated from its path due to its interaction with the particles of the medium through which it passes.”

Daily life phenomena based on scattering of light :

- Blue colour of the sky.
- Reddishness at sunset and sunrise.
- Clouds appear white.
- Danger signals are red.

Question 9.

Define power of the lens. Write its unit. For two thin lenses in combination, derive the relation \(\frac{1}{F}=\frac{1}{f_{1}}+\frac{1}{f_{2}}\)

Answer:

Power of a lens can be defined as the deviation produced by the lens in the rays coming parallel to principal axis at a unit distance from the principal axis.

Unit : Dioptre (D)

Combination of Thin Lenses : Two lenses L_{1}; L_{2} of respective focal lengths f_{1} and f_{2} are kept in contact (figure) A point object O is situated at a distance u in front of the combination and the final image is formed at I. The intermediate image I’ formed by first lens, behaves as virtual image for the second lens.

From the lens formula for first lens,

### RBSE Class 12 Physics Chapter 11 Long Answer Type Questions

Question 1.

Define spherical mirror. Derive the formula connecting object distance, image distance and focal length for the mirror.

Answer:

If some portion of a hollow sphere is cut and polished on one surface, it behaves like mirror i.e., it starts the reflection of light. This mirror is known as spherical mirror. Spherical mirrors are of two types :

(i) Concave mirror

(ii) Convex mirror

(i) Concave Mirror : When the convex surface of spherical glass plate is polished, then the obtained mirror is known as concave mirror (fig 11.3).

(ii) Convex Mirror : When the concave side of the spherical glass plate is polished, then the obtained mirror is known as convex mirror (fig. 11.4).

Mirror Formula or Mirror Equation

The relation among focal length (f), the distance of the object (u) from the mirror and the distance of the image (v) from the mirror, is known as ‘Mirror Formula’ which is given below :

\(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)

Mirror Formula for Convex Mirror : M_{1}M_{2} is a convex mirror. Image of the object AB kept in front of the mirror is A’ B’ behind the mirror. EN is the normal on principal axis from the incident point E of the ray parallel to the principal axis.

Now for ∆ABC and ∆A’B’C will be identical. In these similar triangles

\(\frac{A B}{A^{\prime} B^{\prime}}=\frac{A C}{A^{\prime} C}\) …………….. (1)

Similarly ∆ENF and ∆A’ B’ F will also be similar triangles. Hence in these identical triangles :

\(\frac{E N}{A^{\prime} B^{\prime}}=\frac{N F}{A^{\prime} F}\) …………… (2)

If the aperture of the mirror is small, then point N will be closer to pole P. Therefore, NF ~ PF can be taken. Hence,

\(\frac{A B}{A^{\prime} B^{\prime}}=\frac{P F}{A^{\prime} F}\) ……………… (4)

Now comparing equations (1) and (4), we get,

This formula is also true for concave mirror.

The ratio of the length of the image formed by the mirror (h’) and the length of the object (h) is known as the ‘Magnification’. It is denoted by ‘ m’.

m = \(\frac{h^{\prime}}{h}\)

Question 2.

Write the formation of image by convex lens and concave lens in different situations. Draw relevant diagrams.

Answer:

Image Formation by Lens

For making the image by thin lens, the ray diagram is traced according to following three rules.

(i) The rays passing through optical centre emerge from the lens without deviation.

(ii) The ray coming parallel to the principal axis, after refraction from the lens, meet at focus in convex lens and or appear to diverge from the focus in concave lens, (figure 11.44)

(iii) The rays passing through the focus of convex lens or coming toward the focus of concave lens, become parallel to the principal axis after refraction through the lens (figure 11.38. )

Question 3.

Write the types of lenses. Using the diagram, derive the relation between object distance, image distance and focal length a lens.

Answer:

“The medium is bound by two transparent and curved homogenous medium atleast one surface is called the lens.” Curved surface may be spherical, cylindrical or parabolic. Generally the curved surface is spherical. Mainly the lenses are of two types :

(1) Convex lens and

(2) Concave lens.

1. Convex lens : The lenses which are thinner at the edges and thicker at the centre, are called convex lenses. It converges a parallel beam of light on refraction through it. It has a real focus.

Types of Convex Lenses :

(i) Double convex or biconvex lens : In this lens, both surfaces are convex [fig. 11.27(a)],

(ii) Plano-convex lens : In this lens, one surface is convex and other surface is plane [fig. 11.27(b)],

(iii) Concavo-convex lens : In this lens, one surface is concave and the other surface is convex [fig. 11.27(c)],

2. Concave lens : It is thinner at the centre and thicker at the edges. It diverges a parallel beam of light on refraction through it. It has a virtual focus.

Types of Concave Lenses :

(i) Double concave or biconcave lens : In this lens, both sides are concave [fig. 11.28(a)].

(ii) Plano-concave lens : In this lens, one side is plane and the other is concave [fig. 11.28(b)],

(iii) Convexo-concave lens : In this lens, one side is convex and the other is concave [fig. 11.28(c)].

Converging action of convex lens and diverging action of concave lens :

Convex lens bends the parallel rays incident on it towards the principal axis and centralises them at a point, therefore it is called ‘Convergent Lens’ also, on the other hand the concave lens shifts the parallel rays incident on it away from the principal axis i.e., spreads them, therefore it is called ‘Divergent Lens’ also.

When the medium on both sides of the lens is the same : In figure 11.30 a thin lens is shown on both side of which there is the same medium air. The radii of curvature of both the surfaces of the lens are R_{1} and R_{2} respectively. The image of a point object O, situated at the principal axis, is formed as I’ by refraction through surface AP_{1}B and the image of virtual object I’ is formed as I, the final image by refraction through second surface AP_{2}B of the lens.

The image I’ formed by first surface behaves as virtual object for second surface and the final image is formed as I, therefore for refraction through second surface of the lens,

because the distance of J’ from the second surface is (v’ – t) where t is the thickness of the lens.

This is called ‘Lens Maker’s Formula’.

In general, the equation (4) can be written as under,

This is known as lens maker formula.

If the lens is placed in a liquid e.g., in water, then the focal length of lens in water f_{w} will be obtained by,

Question 4.

Derive the relation between u, v and R for a convex spherical surface, when ray of light is going from rarer to denser medium. Draw appropriate diagram.

Answer:

Refraction through Spherical Surface

Refraction through spherical surfaces also obey the same laws as for plane surfaces.

Sign convention of co-ordinate geometry for measuring the distances from spherical surfaces :

1. The rays of light are always incident on spherical surface from left side.

2. For spherical surface all the distances are measured from pole along the principal axis.

3. The distances measured in the direction of incident ray are taken as positive.

4. The distances measured in opposite direction of incident ray are taken as negative.

5. The lengths of object and image are taken as positive on the upper side of principal axis and negative on the lower side of principal axis.

Formula for refraction of convex spherical surface : Suppose AB is a convex spherical surface, on the left side of which there is one medium (rarer) and on the right side the second medium (denser). P is the pole of the spherical surface and C is its centre of curvature. According to ray diagram of fig. 11.26, I is the image of the object O. The normal drawn from point M on principal axis is MP’.

According to Snell’s law,

_{1}n_{2} = \(\frac{\sin i}{\sin r}\)

Where _{1}n_{2} = the refractive index of medium (2) with respect to medium (1).

n = \(\frac{\sin i}{\sin r}\)

(on taking n in place of _{1}n_{2} temporarily) If the angles i and r are very small, then

sin i ≈ i and sinr ≈ r

∴ n = \(\frac{i}{r}\)

or i = nr …………. (1)

∴ In triangle, the exterior angle is equal to sum of the opposite two interior angles.

∴ From ∆MOC, i = α + γ ………. (2)

And from ∆MIC,

r = β + γ ………… (3)

Substituting the value of i and r in equation (1), then we get,

(α + γ) = n (β + γ) ……….. (4)

If point M is not far from the principal axis, then

(i) Points P and P’ are very close to each other and considered to be same point.

(ii) Angles α, β and γ will be small.

Now, substituting the values of α, β and γ in equation (4) then, we get

This formula is known as ‘Refraction Formula’ of convex surface. If the absolute refractive indices of medium 1 and 2 be n_{1} and n_{2} respectively, then

Question 5.

Draw a labelled diagram to illustrate the action of compound microscope for near point of the eye.

Answer:

Compound Microscope

Compound microscope is an optical instrument which is used to see highly magnified images of very fine particles and very tiny objects.

Construction : It consists of two convex lenses. One lens which is of short focal length and small aperture and remains towards the object, is called ‘object lens’ or ‘field lens’. Second lens which is of larger focal length and aperture and remains towards eye is called ‘eye lens’. Both the lenses are fitted coaxially at both ends of a tube and the distance between these lenses can be changed by rack and pinion method.

Adjustment and Formation of Image : In adjustment process, eye lens or eye pieces is adjusted first. For this, eye piece is moved forward or backward upto such extent that cross wire should be seen clearly. Now the object is placed infront of field lens and it is moved to adjust in such a way that the clear image of the object would be seen. In this situation inverted, larger and virtual image of the object forms on cross-wire. The ray diagram of formation of image is given in figure 11.65.

AB is a small object and its inverted, real and larger image A’ B’ is formed by objective lens. This image acts as virtual object for eye lens, therefore it is moved forward or backward such that the image A’ B’ may lie within the focal length of the eye piece. The image A’ B’ acts as an object for the eye piece which essentially acts like a simple microscope. The eye piece forms a virtual and magnified final image A”B” of the object AB. Clearly, the final image A”B” is inverted with respect to the object A

Magnifying Power : The magnifying power of compound microscope is defined as under ;

(ii) If final image is formed at infinity : The final image will form at infinity only when the image A’ B’ formed by objective lens be at the first focus F’ of eye lens. Therefore,

Similarly, the intermediate image A’ B’ is very near to eye lens. Therefore,

v_{0} = OA’ ≈ OO’ ≈ Length of microscope (L)

or v_{0} ≈ L

From equation (3), the magnifying power,

Question 6.

Draw a diagram to show refraction by prism for a monochromatic light source. Derive relation between angle of prism and minimum deviation angle in terms of refractive index of glass.

Answer:

Refraction in Prism

When a light-ray refracts through the prism, following two actions are possible ;

1. Deviation

2. Dispersion

1. Deviation by Prism: When a monochromatic ray of light incidents on the refracting surface of the prism, then the ray refracts two times and hence the direction of the ray gets deviated (fig. 11.47). “The angle between the directions of incident ray and emergent ray is known as the angle of deviation.”

In diagram it is shown by δ.

In fig. 11.47,

PQ = incident ray; QR = refracted ray; RS = emergent ray; i_{1} = angle of incidence; r_{1} and r_{2} = angle of refraction; i_{2} or e = emergent angle; δ = angle of deviation

Relation between angle of incidence, angle of deviation and angle of prism

In ∆O’QR,

Exterior angle = δ

Angle of Minimum Deviation (δ_{m}) : If the graph is plotted between different incident angles and corresponding angles of deviation, the obtained curve will be as shown in fig. 11.47. It is clear from the curve that angle of deviation first decreases with increasing angle of incidence and reaches a minimum value 8m and then increases. Clearly, any given value of 8 corresponds to two angles of incidence i_{1} and i_{2}. This fact is expected from the symmetry of i_{1} and i_{2} in equation :

i.e., δ remains the same as i_{1} and i_{2} are interchanged. Physically it means that the path of the ray in fig. 11.48 can be traced back resulting the same angle of deviation.

“The minimum value of the angle of deviation suffered by a ray on passing through a prism is called the angle of minimum deviation and is denoted by δ_{m}.”

When i_{1} = i_{2}, then r_{1} = r_{2}.

Hence in the state of minimum deviation :

(i) Angle of incidence is equal to angle of emergence.

(ii) The refracted ray is parallel to the base of the prism.

Formula for Refractive Index of the Prism :

Since in state minimum deviation,

i_{1} = i_{2} = i and r_{1} = r_{2} = r,

The ray diagram through prism is shown in fig. 11.49.

According to Snell’s law, the refractive index of the substance of the prism.

Question 7.

Assuming lens to be two spherical surface, derive relation between u, v and f.

Answer:

When the medium on both sides of the lens is the same : In figure 11.30 a thin lens is shown on both side of which there is the same medium air. The radii of curvature of both the surfaces of the lens are R_{1} and R_{2} respectively. The image of a point object O, situated at the principal axis, is formed as I’ by refraction through surface AP_{1}B and the image of virtual object I’ is formed as I, the final image by refraction through second surface AP_{2}B of the lens.

The image I’ formed by first surface behaves as virtual object for second surface and the final image is formed as I, therefore for refraction through second surface of the lens,

because the distance of J’ from the second surface is (v’ – t) where t is the thickness of the lens.

This is called ‘Lens Maker’s Formula’.

In general, the equation (4) can be written as under,

This is known as lens maker formula.

If the lens is placed in a liquid e.g., in water, then the focal length of lens in water f_{w} will be obtained by,

Question 8.

Write the types of telescope. For a refractive telescope, derive relation for magnifying power. Also discuss about its construction and working.

Answer:

Astronomical Telescope

Telescope is an optical instrument which is used to see the distant object clearly which are not clearly visible by naked eye. Astronomical telescope is used by Astronomic Scientists to see very distant object such as Sky Object, Stars etc. On the basis of construction, the telescope is of two types :

1. Refracting telescope and

2. Reflecting telescope

1. Refracting Type Telescope

Construction : There are two lenses in this telescope. One lens which of larger aperture and focal length and remains towards the object, is called ‘objective lens’ or ‘field lens’.

The other lens which is of short focal length and aperture and remains towards eye, is called ‘eye lens’. Both the lenses are fixed at both ends of a tube and distance between these lenses can be changed by rack and pinion method.

Adjustment and Ray Diagram : First of all eye piece is focussed at cross-wire by moving it forward or backward. Now the objective lens is directed towards the object and it is moved forward or backward till this image of the object is obtained at the cross-wire. In this position the object becomes visible.

Formation of image is shown in figure 11.67 by ray diagram.

The parallel rays coming from the object, situated at infinity, form the inverted, small arid real image A’ B’ at the second focus F_{0} after refraction through objective lens. This intermediate image A’B’ acts as virtual object for eye lens. Hence eye lens is displaced forward or backward till this image comes within first focal plane of eye lens. Thus virtual, larger and erect image A”B” with respect A’B’ is formed which is the final image.

Magnifying power : Magnifying power of telescope is defined as under :

In this position the telescope is said to be in position of “near point adjustment”.

(ii) When the final image forms at infinity : Final image will form at infinity only when the intermediate image A’ B would be at first focus F’e of eye lens. Therefore,

u_{e} = f_{e}

∴ From equation (1), we get,

m = \(-\frac{f_{o}}{f_{e}}\) …………… (3)

In this situation the ray diagram will be as shown in figure 11.68.

In this situation the telescope is said to be in the position of “Normal Adjustment”

### RBSE Class 12 Physics Chapter 11 Numerical Questions

Question 1.

An object is placed at 36 cm in front of concave mirror of focal length 24 cm. Find out the distance of image formed.

Solution:

Given, f = -24 cm, u = -36 cm, v = ?

∴ Distance of image = 72 cm away from mirror towards object.

Question 2.

The relative refractive index of any medium in vacuum is 1.33. The velocity of light in vacuum is c = 3 × 10^{-8} m/s. Then find out velocity of light in medium.

Solution:

Question 3.

The radii of curvatures for a convex lens of a focal length 20 cm are 18 cm and 24 cm. What is the refractive index of the glass?

Solution:

Given : f = +20 cm; R_{1} = +18 cm; R_{2} = -24 cm; n = ?

∴ For thin lens

Question 4.

A ray of light is incident on a glass plate at 50°. If angle of refraction is 30°, then find out refractive index of glass.

Solution:

Given, i = 50° ; r = 30°

Question 5.

An object is at 0.06 m from a convex lens of focal length 0.10 m. Find out the position of the image.

Solution:

Given, f = +0.10; m = + 10 cm; u = -0.06 m = -6 cm; v = ?

By lens formula,

Image will be formed at 15 cm from mirror towards object

Question 6.

The refractive indices of a crown glass prism of 6° refracting angle for red and violet colour are 1.514 and 1.523. Find out angular dispersion produced by prism.

Sol. Given, A = 6° ; n_{R} = 1.514; n_{V} = 1.523; θ = ?

∵ Dispersion for violet and red colour,

θ = A (n_{V} – n_{R} )

∴ θ = 6° (1.523 – 1.514)

= 6° × 0.009

or θ = 0.054°

Question 7.

Two lenses of power +5 D and -7 D are in contact with each other. Find out the power of the combination of the lenses. Combined lens will be convex or concave?

Solution:

Given, P_{1} = +5 D; P_{2} = -7 D; P = ?

∴ The power of combination of the lenses

∴ P = P_{1} + P_{2}

= +5 D – 7 D

or P = – 2D

∵ As power is negative, so combination will be diverging or concave.

Question 8.

The focal length of objective lens and eyepiece for a compound microscope are 0.95 cm and 5 cm, and they are placed at 20 cm from each other. Final image is formed at 25 cm in eyepiece. Find out the magnifying power of the microscope.

Solution:

Given, f_{0} = 0.95 cm; f_{e} = 5 cm; \(\left|v_{o}+u_{e}\right|\) = 20 cm; v_{e} = -25 cm; m = ?

Question 9.

The power of a thin convex lens (n_{g} = 1.5) is +5.0D, when this lens is dipped in a liquid of refractive index n_{l}, it behaves like a concave lens whose focal length is 100 cm, then what will be the value of n_{l}?

Solution:

Question 10.

The refracting angle of a prism is A. The refractive index of the prism is \(\cot \frac{A}{2}\). Find angle of minimum deviation.

Solution:

Given, refractive index of prism n = \(\cot \left(\frac{A}{2}\right)\)

δ_{m} = ?

## Leave a Reply