Question

A circular beam of light of diameter $d=2cm$ falls on a plane surface of glass. The angle of incident is ${60}^o$ and refractive index of glass is $\mu =3/2$. The diameter of the refracted beam is

• A. $2.00cm$
• B. $1.50cm$
• C. $1.63cm$
• D. $2.52cm$

Answer 1

Answer: (C)

$1.63cm$

Using Snell's law:

$\sin \left(i\right)=\mu \sin \left(r\right)\Rightarrow \sin \left(60\right)=\frac{3}{2}\times \sin \left(r\right)\Rightarrow r=\arcsin [\frac{2}{3}\times \sin \left(60\right)]\Rightarrow r=\arcsin \left(\frac{1}{\sqrt{3}}\right)\Rightarrow \cos \left(r\right)=\sqrt{\frac{2}{3}}$

Relation for the diameter of the beam:

$\frac{D_i}{\cos \left(i\right)}=\frac{D_r}{\cos \left(r\right)}\Rightarrow D_r=2\sqrt{\frac{2}{3}}\Rightarrow D_r=1.63cm$

therefore, diameter of the redfracted beam is $1.63cm$