Question

A beam of parallel rays is incident on a transparent slab of refractive index $\sqrt{3}$ making an angle ${30}^{\circ }$ with the surface of the slab. If the width of incident beam of light is 1.732 mm, the width of refracted beam is:

• A. 1.00 mm
• B. 1.50mm
• C. 2.50 mm
• D. 3.00 mm

Answer 1

The correct option is D 3.00 mm

Using ${\mu }_{1}sini={\mu }_{2}sinr$
$1\times sin{60}^{\circ }=\sqrt{3}sinr$

$sinr=\frac{1}{2}$

$AB=\frac{1.732mm}{cosi}$

Now,

$B{B}^1=ABcosr$

$=\frac{1.732}{cosi}\times cosr$

$=\frac{1.732}{\frac{1}{2}}\times \frac{\sqrt{3}}{2}=3mm$