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The angle of incidence for a ray of light at a refracting surface of a prism is 45°. The angle of prism is 60°. If the ray suffers minimum deviation through the prism, the angle of minimum deviation and refractive index of the material of the prism respectively, are
A
30°; $\sqrt{2}$
B
45°; $\sqrt{2}$
C
30°; $\frac{1}{\sqrt{2}}$
D
45°; $\frac{1}{\sqrt{2}}$
Explanation
At minimum deviation i = e; then use the prism formula.
Detailed Solution
For minimum deviation, i = e = 45°, so $\delta_m = i + e - A = 45^\circ + 45^\circ - 60^\circ = 30^\circ$
$\mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} = \frac{\sin45^\circ}{\sin30^\circ} = \frac{1/\sqrt{2}}{1/2}$
$\mu = \sqrt{2}$
$\mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} = \frac{\sin45^\circ}{\sin30^\circ} = \frac{1/\sqrt{2}}{1/2}$
$\mu = \sqrt{2}$