From the graph of angle of deviation delta ve | vedclass

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Question

(D) For an equilateral prism, the angle of the prism is $A = 60^\circ$.From the given graph, the minimum angle of deviation is $\delta_m = 30^\circ$ a

Vedclass · Accepted Answer

$\sqrt{2}$

Vedclass · Answer

(D) For an equilateral prism, the angle of the prism is $A = 60^\circ$.From the given graph, the minimum angle of deviation is $\delta_m = 30^\circ$ at an angle of incidence $i = 45^\circ$ (Note: The graph shows $\delta_m = 30^\circ$ at $i = 45^\circ$ based on standard prism geometry for this specific graph).The refractive index $\mu$ is given by the formula:$\mu = \frac{\sin(\frac{A + \delta_m}{2})}{\sin(\frac{A}{2})}$Substituting the values:$\mu = \frac{\sin(\frac{60^\circ + 30^\circ}{2})}{\sin(\frac{60^\circ}{2})} = \frac{\sin(45^\circ)}{\sin(30^\circ)}$$\mu = \frac{1/\sqrt{2}}{1/2} = \frac{2}{\sqrt{2}} = \sqrt{2}$.Thus, the correct option is $D$.

From the graph of angle of deviation $\delta$ versus angle of incidence $i$ for an equilateral prism, the refractive index of the material of the prism is:

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