The angle of deviation (delta) vs angle of in | vedclass

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(D) From the graph,the minimum deviation $\delta_m = 60^{\circ}$ occurs at an angle of incidence $i = 60^{\circ}$.For minimum deviation,$i = e$,so $\d

Vedclass · Accepted Answer

All the above.

Vedclass · Answer

(D) From the graph,the minimum deviation $\delta_m = 60^{\circ}$ occurs at an angle of incidence $i = 60^{\circ}$.For minimum deviation,$i = e$,so $\delta_m = 2i - A$.$60^{\circ} = 2(60^{\circ}) - A \implies A = 60^{\circ}$. Thus,statement $(A)$ is correct.The refractive index $\mu$ is given by $\mu = \frac{\sin((A + \delta_m)/2)}{\sin(A/2)} = \frac{\sin((60^{\circ} + 60^{\circ})/2)}{\sin(60^{\circ}/2)} = \frac{\sin(60^{\circ})}{\sin(30^{\circ})} = \frac{\sqrt{3}/2}{1/2} = \sqrt{3}$. Thus,statement $(B)$ is correct.For any deviation $\delta$,the relation is $\delta = i + e - A$. From the graph,when $\delta = 65^{\circ}$,the corresponding angle of emergence is $e = 70^{\circ}$ (due to the symmetry of the curve,if $i=70^{\circ}$ gives $\delta=65^{\circ}$,then $i=i_1$ must give $e=70^{\circ}$).$65^{\circ} = i_1 + 70^{\circ} - 60^{\circ} \implies i_1 = 55^{\circ}$. Thus,statement $(C)$ is correct.Therefore,all the statements are correct.

The angle of deviation $(\delta)$ vs angle of incidence $(i)$ is plotted for a prism. Pick the correct statement.

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