The curve of angle of incidence versus angle | vedclass

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(C) From the given graph,the angle of minimum deviation is $\delta_m = 30^{\circ}$ at an angle of incidence $i = 45^{\circ}$. However,looking at the g

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$\frac{\sqrt{3}}{\sqrt{2}}$

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(C) From the given graph,the angle of minimum deviation is $\delta_m = 30^{\circ}$ at an angle of incidence $i = 45^{\circ}$. However,looking at the graph,the minimum deviation occurs at $i = 60^{\circ}$ where $\delta_m = 30^{\circ}$.Wait,re-evaluating the graph: The minimum point is at $i = 60^{\circ}$ and $\delta_m = 30^{\circ}$.For a prism,$\delta_m = 2i - A$. At minimum deviation,$i = e$,so $r_1 = r_2 = r = A/2$.Given $\delta_m = 30^{\circ}$ and $i = 60^{\circ}$,we have $30^{\circ} = 2(60^{\circ}) - A$,which gives $A = 120^{\circ} - 30^{\circ} = 90^{\circ}$.Then $r = A/2 = 45^{\circ}$.The refractive index $\mu$ is given by $\mu = \frac{\sin(i)}{\sin(r)} = \frac{\sin(60^{\circ})}{\sin(45^{\circ})} = \frac{\sqrt{3}/2}{1/\sqrt{2}} = \frac{\sqrt{3}}{\sqrt{2}}$.

The curve of angle of incidence versus angle of deviation shown has been plotted for a prism. The value of the refractive index of the prism used is:

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