Question

A equiangular glass prism of refractive index 1.6 is kept fully immersed in water of refractive index 4/3, for a certain ray of monochromatic light.What is the closest value for the angle of minimum deviation of the light ray in this setup? (Take sine 37^o = 0.6)

• A. 10°
• B. 14°
• C. 18°
• D. 22°

Answer 1

Answer: (B)

Here, angle of prism, $A = 60°$
Refractive index of prism, $\mu_p = 1.6$
Refractive index of water, $\mu_w = \frac{4}{3}$
Angle of minimum deviation, $\delta_m = ?$
As $\frac{\mu_{p}}{\mu_{W}}= \frac{\sin\left(\frac{A+\delta_{m}}{2}\right)}{\sin\left(\frac{A}{2}\right)}$
$ \frac{1.6}{\frac{4}{3}} = \frac{\sin\left(\frac{60^\circ +\delta_{m}}{2}\right)}{\sin\left(\frac{60^\circ}{2}\right)}$
$1.2 = 2 \, \sin\left(30^\circ \frac{\delta_{m}}{2}\right)$
$\sin \, 37^\circ = \sin\left(30^\circ \frac{\delta_{m}}{2}\right)$
$37^\circ = 30^\circ + \frac{\delta_{m}}{2} ; \frac{\delta_{m}}{2} = 7^\circ$
$\therefore \:\: \delta_m = 14^\circ$