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 $\frac{{\mu }_p}{{\mu }_W}=\frac{\sin \left(\frac{A+{\delta }_m}{2}\right)}{\sin \left(\frac{A}{2}\right)}$
$\frac{1.6}{\frac{4}{3}}=\frac{\sin \left(\frac{60^{\circ }+{\delta }_m}{2}\right)}{\sin \left(\frac{60^{\circ }}{2}\right)}$
$1.2=2\sin \left(30^{\circ }\frac{{\delta }_m}{2}\right)$
$\sin 37^{\circ }=\sin \left(30^{\circ }\frac{{\delta }_m}{2}\right)$
$37^{\circ }=30^{\circ }+\frac{{\delta }_m}{2};\frac{{\delta }_m}{2}=7^{\circ }$
$\therefore {\delta }_m=14^{\circ }$