A light ray is incident at an angle theta on | vedclass

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Question

(A) For a prism,the relation between the angle of incidence $i$,the angle of emergence $e$,the prism angle $A$,and the angle of deviation $\delta$ is

Vedclass · Accepted Answer

$7.5$

Vedclass · Answer

(A) For a prism,the relation between the angle of incidence $i$,the angle of emergence $e$,the prism angle $A$,and the angle of deviation $\delta$ is given by $i + e = A + \delta$. Given that the ray emerges normally from the other surface,the angle of emergence $e = 0^{\circ}$. For a prism,the angle of refraction at the first surface $r_1$ and the angle of incidence at the second surface $r_2$ are related to the prism angle by $r_1 + r_2 = A$. Since the ray emerges normally,$r_2 = 0^{\circ}$,which implies $r_1 = A = 5^{\circ}$. Using Snell's law at the first surface: $\mu = \frac{\sin i}{\sin r_1}$. For small angles,$\sin i \approx i$ and $\sin r_1 \approx r_1$. Therefore,$i = \mu 	imes r_1 = 1.5 	imes 5^{\circ} = 7.5^{\circ}$.

$A$ light ray is incident at an angle $\theta$ on the refracting surface of a prism and emerges normally from the other surface. If the prism angle is $5^{\circ}$ and the refractive index of the material of the prism is $1.5$,then the angle of incidence is ...$^{\circ}$.

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