A prism (mu=1.5) has a refracting angle of 30 | vedclass

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Question

(A) When a ray is incident normally on one surface of the prism,it passes undeviated into the prism. Let the prism be $ABC$ with refracting angle $A =

Vedclass · Accepted Answer

$18^{\circ} 36^{\prime}$

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(A) When a ray is incident normally on one surface of the prism,it passes undeviated into the prism. Let the prism be $ABC$ with refracting angle $A = 30^{\circ}$.At the second surface $AB$,the angle of incidence $i$ is equal to the refracting angle of the prism,so $i = 30^{\circ}$.Using Snell's law at the second surface: $\mu \sin i = 1 \sin e$,where $e$ is the angle of emergence.$1.5 	imes \sin 30^{\circ} = \sin e$$1.5 	imes 0.5 = \sin e$$\sin e = 0.75$Given $\sin 48^{\circ} 36^{\prime} = 0.75$,therefore $e = 48^{\circ} 36^{\prime}$.The angle of deviation $\delta$ for a ray incident normally is given by $\delta = e - i$.$\delta = 48^{\circ} 36^{\prime} - 30^{\circ} = 18^{\circ} 36^{\prime}$.

$A$ prism $(\mu=1.5)$ has a refracting angle of $30^{\circ}$. The deviation of a monochromatic ray incident normally on one of its surfaces will be (given,$\sin 48^{\circ} 36^{\prime}=0.75$):

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