If the refractive index of the material of an equilateral prism is $\sqrt{2}$, then the angle of min

Question

If the refractive index of the material of an equilateral prism is $\sqrt{2}$, then the angle of minimum deviation for the prism is

Professor PC Thomas Classes · Answer

(c) : For an equilateral prism, the angle of the prism is $A = 60^\circ$. The refractive index is given as $\mu = \sqrt{2}$. The formula relating refractive index and minimum deviation is $\mu = \sin((A + \delta_m) / 2) / \sin(A / 2)$. Substituting the values: $\sqrt{2} = \sin((60^\circ + \delta_m) / 2) / \sin(30^\circ)$. Since $\sin(30^\circ) = 1/2$, we get $\sqrt{2} 	imes (1/2) = \sin((60^\circ + \delta_m) / 2)$, which simplifies to $1/\sqrt{2} = \sin((60^\circ + \delta_m) / 2)$. Because $\sin(45^\circ) = 1/\sqrt{2}$, we have $(60^\circ + \delta_m) / 2 = 45^\circ$. Solving for $\delta_m$ yields $60^\circ + \delta_m = 90^\circ$, so $\delta_m = 30^\circ$.