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 $1/\sqrt{3}$ m, the angle of refraction produced is

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

Hint:

$\sin(i-r) = \sin i \cos r - \cos i \sin r$.

Answer 1

The Correct Option is B

Step 1: Formula
Lateral shift $d = \frac{t \sin(i-r)}{\cos r}$. Given $\frac{d}{t} = \frac{1}{\sqrt{3}}$ and $i = 45^{\circ}$.
Step 2: Expansion
$\frac{1}{\sqrt{3}} = \frac{\sin 45 \cos r - \cos 45 \sin r}{\cos r} = \frac{1}{\sqrt{2}}(1 - \tan r)$.
Step 3: Solve
$1 - \tan r = \frac{\sqrt{2}}{\sqrt{3}}$, so $\tan r = 1 - \frac{\sqrt{2}}{\sqrt{3}}$.
Final Answer: (b)