The angle of a prism is $A$. One of its refracting surfaces is silvered. Light rays falling at an angle of inc

Question

The angle of a prism is $A$. One of its refracting surfaces is silvered. Light rays falling at an angle of incidence $2A$ on the first surface returns back through the same path after suffering reflection at the silvered surface. The refractive index $\mu$ of the prism is:

Accepted Answer

1.  **Condition for Retracing Path:** For a light ray to return back through the same path after reflection from a plane mirror (the silvered surface), it must strike the mirror normally. This implies that the angle of incidence at the second surface is zero.
2.  **Prism Geometry:** For a prism with angle $A$ and refracting angle $r_1$ at the first surface and $r_2$ at the second surface, the relation is $r_1 + r_2 = A$. Since the ray strikes the second surface normally, $r_2 = 0$. Therefore, $r_1 = A$.
3.  **Snell's Law at First Surface:**
    *   Angle of incidence, $i = 2A$ (Given).
    *   Angle of refraction, $r_1 = A$.
    *   Using Snell's Law: $\mu_1 \sin i = \mu_2 \sin r_1$.
    *   Assuming the surrounding medium is air ($\mu_1 = 1$), we have: $1 \cdot \sin(2A) = \mu \sin(A)$.
4.  **Trigonometric Identity:**
    *   $\sin(2A) = 2 \sin A \cos A$.
5.  **Calculation:**
    *   $2 \sin A \cos A = \mu \sin A$
    *   $\mu = 2 \cos A$.

Step-by-Step Solution

Practice All PHYSICS Questions