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(A) Given: Prism angle $A = 30^{\circ}$,angle of incidence $i_1 = 60^{\circ}$,and angle of deviation $\delta = 30^{\circ}$.For a prism,the deviation f

Vedclass · Accepted Answer

$1.732$

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(A) Given: Prism angle $A = 30^{\circ}$,angle of incidence $i_1 = 60^{\circ}$,and angle of deviation $\delta = 30^{\circ}$.For a prism,the deviation formula is $\delta = (i_1 + i_2) - A$.Substituting the values: $30^{\circ} = (60^{\circ} + i_2) - 30^{\circ}$.$30^{\circ} = 30^{\circ} + i_2$,which gives $i_2 = 0^{\circ}$.Since the emergent angle $i_2 = 0^{\circ}$,the emergent ray is normal to the second face,implying the angle of refraction at the second face $r_2 = 0^{\circ}$.Using the relation $r_1 + r_2 = A$,we get $r_1 + 0^{\circ} = 30^{\circ}$,so $r_1 = 30^{\circ}$.Applying Snell's law at the first face: $\mu_1 \sin i_1 = \mu_2 \sin r_1$.Assuming the surrounding medium is air $(\mu_1 = 1)$: $1 \cdot \sin 60^{\circ} = \mu_2 \sin 30^{\circ}$.$\frac{\sqrt{3}}{2} = \mu_2 \cdot \frac{1}{2}$.Therefore,$\mu_2 = \sqrt{3} \approx 1.732$.

$A$ ray of light is incident at $60^{\circ}$ on one face of a prism of angle $30^{\circ}$ and the emergent ray makes $30^{\circ}$ with the incident ray. The refractive index of the prism is $(\sin 30^{\circ}=0.5, \sin 60^{\circ}=\sqrt{3}/2)$.

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