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

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

Vedclass · Accepted Answer

All the above

Vedclass · Answer

(D) From the graph,at minimum deviation,$\delta_m = 60^{\circ}$ and the corresponding angle of incidence $i = 60^{\circ}$.For minimum deviation,$i = e$,so $\delta_m = 2i - A$.$60^{\circ} = 2(60^{\circ}) - A \Rightarrow A = 60^{\circ}$. Thus,option $(A)$ 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,option $(B)$ is correct.For a deviation $\delta = 65^{\circ}$,the graph shows two possible angles of incidence,$i_1$ and $70^{\circ}$. Since $\delta = i + e - A$,for $i = 70^{\circ}$,$65^{\circ} = 70^{\circ} + e - 60^{\circ} \Rightarrow e = 55^{\circ}$. By the principle of reversibility,if $i = 70^{\circ}$ gives $e = 55^{\circ}$,then $i = 55^{\circ}$ gives $e = 70^{\circ}$. Thus,$i_1 = 55^{\circ}$. Option $(C)$ 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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