Question

Figure shows an irregular block of material of refractive index $\sqrt{2}$. A ray of light strikes the face AB as shown in the figure. After refraction it is incident on a spherical surface CD of radius of curvature 0.4 m and enters a medium of refractive index 1.514 to meet PQ at E. The distance OE is

• A. $4m$
• B. $8m$
• C. $6m$
• D. $12m$

Answer 1

Answer: (C)

$6m$

By Snell's law
${\mu }_{1}sini={\mu }_{2}sinr$
From the question,
$1\times sin{45}^{\circ }=\sqrt{2}\times sinr$
Solving this, we get $r={30}^{\circ }$
For this r, the ray after refraction from the first surface will travel parallel to PQ as shown.
For refraction at curved surface CD, it will appear as though $u=-\infty$ as the ray is parallel to principal axis PQ.
Given $R=0.4m$
Putting this and the given values in the formula

$\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}$

we get

$\frac{1.514}{v}-0=\frac{1.514-\sqrt{2}}{0.4}$

Solving this we get $v\approx 6m$