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

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(A) Given: Refractive index $\mu = 1.5$,Prism angle $A = 30^\circ$.Since the ray is incident normally on one surface,the angle of incidence $i_1 = 0^\

Vedclass · Accepted Answer

$18^\circ 36'$

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(A) Given: Refractive index $\mu = 1.5$,Prism angle $A = 30^\circ$.Since the ray is incident normally on one surface,the angle of incidence $i_1 = 0^\circ$,which implies the angle of refraction $r_1 = 0^\circ$.Using the relation $r_1 + r_2 = A$,we get $0^\circ + r_2 = 30^\circ$,so $r_2 = 30^\circ$.Applying Snell's Law at the second surface: $\mu \sin r_2 = 1 \cdot \sin e$,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 $\sin 48^\circ 36' = 0.75$,therefore $e = 48^\circ 36'$.The deviation $\delta$ for a prism is given by $\delta = i_1 + e - A$.Since $i_1 = 0^\circ$,$\delta = e - A = 48^\circ 36' - 30^\circ = 18^\circ 36'$.Thus,the correct option is $A$.

$A$ 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 $(\sin 48^\circ 36' = 0.75)$.

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