Question

A light beam of diameter 'd' is incident on a glass plate of refractive index $\frac{3}{2}$. If $\theta ={60}^{\circ }$ diameter of beam in glass becomes 'D', where $D=\left(a\sqrt{\frac{b}{c}}\right)d$ then $[\frac{c}{a}+b]$ is (a, b and c are integers less than 5. Write upto two digits after the decimal point.)

Answer 1

By Snell's law

$n=\frac{\sin \theta }{\sin r}$

where $\sin (90-\theta )=\frac{d}{AB}$ and $\sin (90-r)=\frac{D}{AB}$

So $\frac{\cos \theta }{\cos r}=\frac{d}{D}\Rightarrow D=d\frac{\cos r}{\cos \theta }=\frac{d}{\cos \theta }\sqrt{1-{\sin }^{2}r}$

$D=\frac{d}{\cos \theta }\sqrt{1-\frac{{\sin }^2\theta }{n^2}}=\frac{d\sqrt{n^2-{\sin }^2\theta }}{ncos\theta }=2\sqrt{\frac{2}{3}}d$