Ncert Exercises & Solutions

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°. 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.

Hint: $\mu=\frac{sin\frac{A+\delta _{m}}{2}}{sin\frac{A}{2}}$
Step 1: Find the refractive index of the prism.
Let the refractive index of the material of the prism = $μ ‘$
The refractive index of the prism is given by:

$μ^{'} = \frac{\sin (\frac{A + δ_{m}}{2})}{\sin \frac{A}{2}}$
$= \frac{\sin (\frac{60^{\circ} + 40^{\circ}}{2})}{\sin \frac{60^{\circ}}{2}} = \frac{\sin 50^{\circ}}{\sin 30^{\circ}} = 1.532$

$Step 2: Find$ the new angle of minimum deviation $δ'_{m}$ be the new angle of minimum deviation for the same prism when it is placed in water.
The refractive index of glass with respect to the water is given by:

$μ_{g w} = \frac{μ^{'}}{μ} = \frac{\sin (\frac{A + {δ_{m}}^{'}}{2})}{\sin \frac{A}{2}}$
$\sin (\frac{A + {δ_{m}}^{'}}{2}) = \frac{μ^{'}}{μ} \sin \frac{A}{2}$
$\sin (\frac{A + {δ_{m}}^{'}}{2}) = \frac{1.532}{1.33} \times \sin \frac{60 °}{2} = 0.5759$

$(\frac{A + {δ_{m}}^{'}}{2}) = \sin^{- 1} 0.5759 = 35.16 °$
$60 ° + δ_{m}' = 70.32 °$
$∴ δ_{m}' = 70.32 ° - 60 ° = 10.32 °$

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

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