A ray of light passing from air through an equilateral glass prism undergoes minimum deviation when the angle of incidence is \displaystyle\dfrac{3}{4} th of the angle of prism. Calculate the speed of light in the prism.

Here, Angle of prism, A=60^o Angle of incidence, i= \dfrac{3}{4}\times 60 i+e=A+ \delta e is emergence anglewhere \delta is deviation angleAt minimum deviation \delta=\delta_{min} i=e2i=A+ \delta_{min} 2\times 45=60+\delta_{min} \delta_{min}=30 By prism law, \mu=\dfrac{\sin\dfrac{A+\delta_{min}}{2}}{\sin\dfrac{A}{2}} \mu=\dfrac{\sin\dfrac{60+30}{2}}{\sin\dfrac{60}{2}}=\dfrac{2}{\sqrt{2}}=1.41 Let v be velocity of light in prism v=\dfrac{c}{\mu}=\dfrac{3\times 10^8}{1.41} = 2\times 10^8 m/s

Ray Optics And Optical Instruments - Difficult Conceptual Numericals

Q: A flat mirror revolves at a constant angular velocity making n=0.4 revolutions per second. With what velocity (in ms^{-1} ) will a light spot move along a spherical screen with a radius of 15 metres, if the mirror is at the centre of curvature of the screen?

\because Angular velocity of mirror = 0.4rev/s \Rightarrow 0.4\times 2\pi=0.8\pi\,rad/s \because Angular velocity of reflected ray \Rightarrow 2\times 0.8\pi=1.6\pi\,rad/s Hence, velocity of light spot over the screen v=rw=15\times 1.6\pi=75.4m/s

Q: A prism having an apex angle of 4^0 and refractive index of 1.50 is located in front of a vertical plane mirror as shown.A horizontal ray of light is incident on the prism. The total angle through which the ray is deviated is:

We have the deviation of prism, \delta _1 = A (\mu -1) = 4^o ( 1.5-1) \implies \delta_1 = 2^o For a plane mirror, i = 2^o \delta_2 = 180 - 2(2^o) = 176^o \implies \delta =\delta_1 + \delta_2 = 176^o+2^o=178^o

Q: 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^o . What is the refractive index of the material of the prism? The refracting angle of the prism is 60^o . If the prism is placed in water (refractive index 1.33), predict the new angle of minimum deviation of a parallel beam of light.

The angle of the prism, 60^o .Minimum deviation \delta=40^o \mu^{'} is the refractive index of the material of the prism. \mu^{'}=\dfrac{sin\left(\dfrac{A+\delta m}{2}\right)}{sin\left(\dfrac{A}{2}\right)}=1.5321 Let \delta_1 be new minimum deviation, now it is put under water \mu=1.33 \Rightarrow\dfrac{ \mu^{'}}{\mu}=\dfrac{sin\left(\dfrac{A+\delta m}{2}\right)}{sin\left(\dfrac{A}{2}\right)} \Rightarrow \delta_1=10.32^o

Q: A compound microscope is used to enlarge an object kept at a distance 0.03\ m from it's objective which consists of several convex lenses in contact and has focal length 0.02\ m . If a lens of focal length 0.1\ m is removed from the objective, then by what distance the eye piece of the microscope must be moved to refocus the image

If initially the objective ( focal length F ) forms the image at distance v_o then v_o =\dfrac{u_of_o}{u_o -f_o}=\dfrac{3\times 2}{3-2}=6\ cm Now as in case of lenses in contact \dfrac {1}{F_0}=\dfrac{1}{f_1}+\dfrac {1}{f_2}+\dfrac{1}{f_3}+....=\dfrac{1}{f_1}+\dfrac{1}{F_0'} \left\{ where\ \dfrac{1}{F_0'}=\dfrac{1}{f_2}+\dfrac{1}{f_3}+....\right\} So if one of the lens is removed, the focal length of the remaining lens system \dfrac {1}{F_0'}=\dfrac{1}{F_0}-\dfrac{1}{f_1}=\dfrac 12-\dfrac {1}{10}\Rightarrow F_0'=2.5\ cm This lens will form the image of same object at a distance v_0' such that v_0'=\dfrac{u_0F_0'}{u_0-F_0'}=\dfrac{3\times 2.5}{(3.-2.5)}=15\ cm So to refocus the image, eye-piece must be moved by the same distance through which the image formed by the objective has shifted i.e. 15-6=9\ cm .

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Difficult Conceptual Numericals

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Ray Optics And Optical Instruments

Rotating plane mirrors:
Questions involving rotation of mirrors can be difficult at times. The very thing to remember is, if the plane mirror rotates by an angle

θ

, then the reflected ray rotates by an angle 2

θ

in the same direction.

A flat mirror revolves at a constant angular velocity making

n = 0.4

revolutions per second. With what velocity (in

m s^{- 1}

) will a light spot move along a spherical screen with a radius of

15

metres, if the mirror is at the centre of curvature of the screen?

37.7

60.3

68.7

75.4

90.4

View solution

Combination of prism and mirror:
The total deviation of light in any combination can simply be found by the algebraic sum of the deviations produced by each optical element. For combination of prism and mirror, deviation produced by prism is added to the deviation produced by the mirror.

A prism having an apex angle of

4^{0}

and refractive index of 1.50 is located in front of a vertical plane mirror as shown.A horizontal ray of light is incident on the prism. The total angle through which the ray is deviated is:

4^{0}

clockwise

17 8^{0}

clockwise

2^{0}

clockwise

8^{0}

clockwise

View solution

Finding refractive index of prism when angle of minimum deviation is given:
Sometimes, the angle of minimum deviation is not given. In fact, we will have to apply simple geometry to find the relation between the angle of prism A and angle of minimum deviation. There can be many layer to the question. Instead of asking the refracting index directly, the examiner might ask the factor by which light has slowed down in that medium, which is obviously, the R.I. of the medium.

A ray of light passing from air through an equilateral glass prism undergoes minimum deviation when the angle of incidence is

\frac{3}{4}

th of the angle of prism. Calculate the speed of light in the prism.

View solution

An optical system immersed in a medium of given refractive index.
Usually, we come across questions where the optical systems are in placed in air. So using the absolute refractive indices for the lenses or prisms involved in place of relative R.I. doesn't make any difference. However we should be careful with equations where the system is immersed in an optical medium other than air.

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

4 0^{o}

. What is the refractive index of the material of the prism? The refracting angle of the prism is

6 0^{o}

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

View solution

Compound microscope which has another combination of lenses:
In such questions, we use the very basic concepts step by step. We first reduce the combination to a single equivalent lens. And then use the results obtained for compound microscope for the final two lenses: the objective and the eyepiece.

A compound microscope is used to enlarge an object kept at a distance

0.03 m

from it's objective which consists of several convex lenses in contact and has focal length

0.02 m

. If a lens of focal length

0.1 m

is removed from the objective, then by what distance the eye piece of the microscope must be moved to refocus the image

2.5 c m

6 c m

15 c m

9 c m

View solution