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(A) For the full prism,the ray is incident normally,so the angle of incidence at the first face is $0^{\circ}$. The angle of refraction at the first f

Vedclass · Accepted Answer

emerge at an angle of emergence $\sin^{-1} \left( \frac{1}{2} \sec \frac{A}{2} \right)$

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(A) For the full prism,the ray is incident normally,so the angle of incidence at the first face is $0^{\circ}$. The angle of refraction at the first face is $r_1 = 0^{\circ}$.Since $r_1 + r_2 = A$,we have $r_2 = A$.For grazing emergence,the angle of emergence $e = 90^{\circ}$. Applying Snell's law at the second face: $\mu \sin r_2 = 1 \cdot \sin e \Rightarrow \mu \sin A = \sin 90^{\circ} = 1$,so $\mu = \frac{1}{\sin A}$.When half the prism is removed,the ray hits the internal face at an angle of incidence $r' = A/2$ (by geometry of the bisected prism).Applying Snell's law at this new interface: $\mu \sin r' = 1 \cdot \sin e'$,where $e'$ is the new angle of emergence.$\frac{1}{\sin A} \cdot \sin \frac{A}{2} = \sin e'$$\sin e' = \frac{\sin(A/2)}{2 \sin(A/2) \cos(A/2)} = \frac{1}{2 \cos(A/2)} = \frac{1}{2} \sec \frac{A}{2}$.Thus,$e' = \sin^{-1} \left( \frac{1}{2} \sec \frac{A}{2} \right)$.

$A$ ray of light is incident normally on the first refracting face of a prism of refracting angle $A$. The ray of light comes out at grazing emergence. If one half of the prism (shaded portion) is knocked off,the same ray will:

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