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

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(B) Let the angle of incidence be $i = 2A$ and the angle of refraction be $r_1$. According to Snell's Law at the first surface: $\mu = \frac{\sin i}{\

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$2 \cos A$

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(B) Let the angle of incidence be $i = 2A$ and the angle of refraction be $r_1$. According to Snell's Law at the first surface: $\mu = \frac{\sin i}{\sin r_1} = \frac{\sin 2A}{\sin r_1}$. Since the light ray returns back through the same path after reflection at the silvered surface,it must strike the silvered surface normally. This implies the angle of refraction at the second surface is $r_2 = 0$. For a prism,$A = r_1 + r_2$. Since $r_2 = 0$,we have $r_1 = A$. Substituting $r_1 = A$ into the Snell's Law equation: $\mu = \frac{\sin 2A}{\sin A}$ Using the trigonometric identity $\sin 2A = 2 \sin A \cos A$: $\mu = \frac{2 \sin A \cos A}{\sin A} = 2 \cos A$.

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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