Question

The angle of minimum deviation in an equilateral prism of refractive index 1.414 is

• A. $60^\circ$
• B. $30^\circ$
• C. $90^\circ$
• D. $45^\circ$

Answer 1

Answer: (B)

$30^\circ$

Answer 2

$\mu = \frac{\bigg(\sin\frac{A+\delta_m}{2}\bigg)}{\sin\frac{A}{2}}$
or $1.414 = \frac{\bigg(\sin\frac{60^\circ+\delta_m}{2}\bigg)}{\sin\frac{60^\circ}{2}}$
$$ (for equilateral prism, $A = 60^\circ$)
or $\sqrt 2 = \frac{\sin\bigg[\frac{60^\circ+\delta_m}{2}\bigg]}{\sin 30^\circ } (\because \sqrt 2 \approx 1.414 )$
$\sqrt 2 \times \frac{1}{2} = \sin\bigg(\frac{60^\circ+\delta_m}{2}\bigg)$
or $\frac{1}{\sqrt 2} = \sin\bigg(\frac{60^\circ+\delta_m}{2}\bigg)$
$sin\, 45^\circ = sin\bigg(\frac{60^\circ+\delta_m}{2}\bigg)$
or $60^\circ + \delta_m = 90^\circ$
or $\delta_m = 30^\circ$