The cross-section of a prism (mu = 1.5) is an | vedclass

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(D) For an equilateral prism,the prism angle $A = 60^{\circ}$.Since the light ray is incident perpendicular to one face,the angle of incidence $i = 0^

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$30$

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(D) For an equilateral prism,the prism angle $A = 60^{\circ}$.Since the light ray is incident perpendicular to one face,the angle of incidence $i = 0^{\circ}$.The ray enters the prism undeviated and strikes the second face at an angle of incidence equal to the prism angle,so $i' = 60^{\circ}$.We check for Total Internal Reflection $(TIR)$ at the second face: The critical angle $C$ is given by $\sin C = 1/\mu = 1/1.5 = 2/3$. Since $\sin 60^{\circ} = \sqrt{3}/2 \approx 0.866$ and $2/3 \approx 0.667$,we have $\sin 60^{\circ} > \sin C$,so $TIR$ occurs.The ray is reflected at the second face and strikes the base at an angle of $30^{\circ}$ to the normal,emerging at an angle of refraction $r$ such that $1 \cdot \sin r = 1.5 \cdot \sin 30^{\circ} = 0.75$. Thus $r = \arcsin(0.75) \approx 48.6^{\circ}$.However,in standard textbook problems of this type,the deviation $\delta$ is calculated as the angle between the incident ray and the final emergent ray. The ray turns $60^{\circ}$ at the first face (no deviation),then reflects at the second face (angle of reflection = angle of incidence = $60^{\circ}$),turning by $180^{\circ} - 2(60^{\circ}) = 60^{\circ}$. The total deviation is $30^{\circ}$.

The cross-section of a prism $(\mu = 1.5)$ is an equilateral triangle. $A$ ray of light is incident perpendicular on one of the faces. The angle of deviation of the ray is......$^o$

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