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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.

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Payal Gupta | Contributor-Level 10

4 months ago

9.6 The angle of minimum deviation, $δ_{m}$ = 40 $°$

Angle of prism, A = 60 $°$

Let the refractive index of water, $= 1.33$ , and the refractive index of prism material = $μ'$

The angle of deviation is related to refractive index $μ'$ is given as

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

So the refractive index of prism material is 1.532

Since the prism is placed in water, let $δ_{m}^{'}$ be the new angle of minimum deviation.

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

$μ_{g}^{w}$ = $\frac{μ^{'}}{μ}$ = $\frac{\sin ? \frac{(A + δ_{m}^{'})}{2}}{\sin ? \frac{A}{2}}$

$\frac{1.532}{1.33}$ = $\frac{\sin ? \frac{(60 ° + δ_{m}^{'})}{2}}{\sin ? \frac{60 °}{2}}$

1.152 $\times \sin ? 30 °$ = $\sin ? \frac{(60 ° + δ_{m}^{'})}{2}$

$\frac{(60 ° + δ_{m}^{'})}{2} = \sin^{- 1} ? 0.576$ = 35.2 $°$

$δ_{m}^{'} = 2 \times$ 35.2 $°$ - 60 $°$ = 10.33&nb

...more

9.6 The angle of minimum deviation, $δ_{m}$ = 40 $°$

Angle of prism, A = 60 $°$

Let the refractive index of water, $= 1.33$ , and the refractive index of prism material = $μ'$

The angle of deviation is related to refractive index $μ'$ is given as

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

So the refractive index of prism material is 1.532

Since the prism is placed in water, let $δ_{m}^{'}$ be the new angle of minimum deviation.

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

$μ_{g}^{w}$ = $\frac{μ^{'}}{μ}$ = $\frac{\sin ? \frac{(A + δ_{m}^{'})}{2}}{\sin ? \frac{A}{2}}$

$\frac{1.532}{1.33}$ = $\frac{\sin ? \frac{(60 ° + δ_{m}^{'})}{2}}{\sin ? \frac{60 °}{2}}$

1.152 $\times \sin ? 30 °$ = $\sin ? \frac{(60 ° + δ_{m}^{'})}{2}$

$\frac{(60 ° + δ_{m}^{'})}{2} = \sin^{- 1} ? 0.576$ = 35.2 $°$

$δ_{m}^{'} = 2 \times$ 35.2 $°$ - 60 $°$ = 10.33 $°$

Hence the new minimum angle of deviation is 10.33 $°$

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