The prism has refracting angle ' $\mathrm{A}$ '. The second refracting surface of the prism is silvered.…

The prism has refracting angle ' $\mathrm{A}$ '. The second refracting surface of the prism is silvered. Light ray falling on first refracting surface with angle of incidence ' $2 \mathrm{~A}$ ', reaches the second surface and returns back through the same path due to reflection at the silvered surface. The refractive index of the material of the prism is
  1. $\frac{1}{2} \sin \mathrm{A}$
  2. $\frac{1}{2} \cos \mathrm{A}$
  3. $2 \sin A$
  4. $2 \cos A$

Solution

Normal incidence at silvered surface $\therefore \quad \mu=\frac{\sin \mathrm{i}}{\sin \mathrm{r}}=\frac{\sin 2 \mathrm{~A}}{\sin \mathrm{A}}=\frac{2 \sin \mathrm{A} \cos \mathrm{A}}{\sin \mathrm{A}}=2 \cos \mathrm{A}$

Asked in: MHT CET 2023 (11 May Shift 1)

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