Question

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

• A. $7m$
• B. $.29m$
• C. $6.06m$
• D. $8.55m$

Answer 1

The correct option is C $6.06m$

From Snell's law,

${\mu }_1\sin i={\mu }_2\sin r$

$\Rightarrow \sin r=\frac{{\mu }_1}{{\mu }_2}\sin i=\frac{1}{\sqrt{2}}\sin {45}^o$

$\Rightarrow \frac{1}{2}=\sin r\Rightarrow r={30}^o$

This means that the ray becomes parallel to side $AD$ inside the slab.

This implies for the second face $CD$, $u=\infty$

Given $R=0.4m$

$\frac{{\mu }_3}{v}-\frac{{\mu }_2}{v}=\frac{{\mu }_3-{\mu }_2}{R}$

$\Rightarrow \frac{1.514}{v}-\frac{\sqrt{2}}{\infty }=\frac{1.514-\sqrt{2}}{R}$

$v=\frac{1.514\times 0.4}{1.514-1.414}=\frac{1.514\times 0.4}{0.1}$

$v=6.056=6.06m$ (upto $2$ decimal places).