The angle of a prism is A. One of its refract | vedclass

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(A) For the light ray to retrace its path after reflection from the silvered surface,it must strike the silvered surface normally (at an angle of $90^

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$2\,sin\,A$

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(A) For the light ray to retrace its path after reflection from the silvered surface,it must strike the silvered surface normally (at an angle of $90^{\circ}$).In the triangle formed by the ray inside the prism,the angle at the silvered surface is $90^{\circ}$ and the prism angle is $A$. Therefore,the angle of refraction $r$ at the first surface must be $r = 90^{\circ} - A$.Applying Snell's law at the first surface:$1 \cdot sin(i) = \mu \cdot sin(r)$Given $i = 2A$ and $r = 90^{\circ} - A$,we have:$sin(2A) = \mu \cdot sin(90^{\circ} - A)$$sin(2A) = \mu \cdot cos(A)$Using the trigonometric identity $sin(2A) = 2\,sin(A)\,cos(A)$:$2\,sin(A)\,cos(A) = \mu \cdot cos(A)$Dividing both sides by $cos(A)$ (assuming $A 
eq 90^{\circ}$):$\mu = 2\,sin(A)$

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 return back through the same path after suffering reflection at the silvered surface. The refractive index $\mu$ of the prism is

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