Question

A ray of light incident on a transparent block at an angle of incidence 60${}^{\circ }$. If the refractive index of the block is 1.732, the angle of deviation of the refracted ray is :

• A. 15${}^{\circ }$
• B. 25${}^{\circ }$
• C. 30${}^{\circ }$
• D. 45${}^{\circ }$

Answer 1

The correct option is C 30${}^{\circ }$

Using Snell's Law :

$\frac{sini}{sinr}=\frac{v_1}{v_2}=\frac{{\mu }_2}{{\mu }_{{}_1}}$

${\mu }_{2}sinr={\mu }_{1}sini$

$1.732\times sinr=1\times sin{60}^{\prime }$

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

$r={30}^{\circ }$

$soangleofdeviation={60}^{\circ }-{30}^{\circ }$

$={30}^{\circ }$