A prism has a refracting angle A. The second | vedclass

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

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$2 \cos A$

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

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