Question

A ray of light is incident on a surface of glass slab at an angle $45^{\circ }.$ If the lateral shift produced per unit thickness is $\frac{1}{\sqrt{3}}m$ , then the angle of refraction produced at the incident surface is

• A. ${\tan }^{-1}\left(\frac{\sqrt{3}}{2}\right)$
• B. ${\tan }^{-1}\left(1-\sqrt{\frac{2}{3}}\right)$
• C. ${\sin }^{-1}\left(1-\sqrt{\frac{2}{3}}\right)$
• D. ${\tan }^{-1}\left(\sqrt{\frac{2}{\sqrt{3}-1}}\right)$

Answer 1

Answer: (B)

Here, angle of incidence $i=45^{\circ }$

$\frac{Lateralshift\left(d\right)}{Thicknessofglassslab\left(t\right)}=\frac{1}{\sqrt{3}}$
Lateral shift $d=\frac{tsin\delta }{cosr}=\frac{tsin(i-r)}{cosr}$
$\Rightarrow \frac{d}{t}=\frac{sin(i-r)}{cosr}$
or $\frac{d}{t}=\frac{sinicosr-cosisinr}{cosr}$
or $\frac{d}{t}=\frac{sin45^{\circ }cosr-cos45^{\circ }sinr}{cosr}=\frac{cosr-sinr}{\sqrt{2}cosr}$
or $\frac{d}{t}=\frac{1}{\sqrt{2}}\left(1-tanr\right)$
or $\frac{1}{\sqrt{3}}=\frac{1}{\sqrt{2}}\left(1-tanr\right)$
or $tanr=1-\frac{\sqrt{2}}{\sqrt{3}}$
oror $r={\tan }^{-1}\left(1-\frac{\sqrt{2}}{\sqrt{3}}\right)$