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Q 9.6 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 $40^{\circ}$ . What is the refractive index of the material of the prism? The refracting angle of the prism is $60^{\circ}$ . If the prism is placed in water, predict the new angle of minimum deviation of a parallel beam of light.

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In Prism :

Prism angle ( $A$ ) = First Refraction Angle ( $r _1$ ) + Second refraction angle ( $r _2$ )

also, Deviation angle ( $\delta$ ) = incident angle( $i$ ) + emerging angle( $e$ ) - Prism angle ( $A$ ) ..............(1)

the deviation angle is minimum when the incident angle( $i$ ) and an emerging angle( $e$ ) are the same. in other words

$i=e$ ...........(2)

from (1) and (2)

$\delta_{min} = 2i-A$

$i=\frac{\delta_{min} +A}{2}$ ..........................(3)

We also have

$r _1 = r_2 = r = \frac{A}{2}$ .................(4)

Now applying snells law using equation (3) and (4)

$\mu _1sini=\mu _2sinr$

$1sin(\frac{\delta _{min}+A}{2})=\mu _2sin\frac{A}{2}$

$\mu _2=\frac{sin(\frac{\delta _{min}+A}{2})}{sin\frac{A}{2}}$ ...................(5)

Given

$\\\delta _{min}= 40 \\A = 60$

putting those values in (5) we get

$\mu _2=\frac{sin(\frac{40+60}{2})}{sin\frac{60}{2}}=\frac{sin50}{sin30}=1.532$

Hence the refractive index of the prism is 1.532.

Now when the prism is in the water.

Applying Snell's law:

$\mu _1sin(\frac{\delta _{min}+A}{2})=\mu _2sin\frac{A}{2}$

$1.33sin(\frac{\delta _{min}+60}{2})=1.532sin\frac{60}{2}$

$sin(\frac{\delta _{min}+60}{2})=\frac{1.532*0.5}{1.33}$

$\frac{\delta _{min}+60}{2}=sin^{-1}\frac{1.532*0.5}{1.33}$

$\delta _{min} =2sin^{-1}0.5759 - 60$

$\delta _{min} =2*35.16- 60 =10.32^0$

Hence minimum angle of deviation inside water is 10.32 degree.

Posted by

Pankaj Sanodiya

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