Question

Two identical glass rods $S_1$ and $S_2$ (refractive index = 1.5) have one convex end of radius of curvature 10 cm. They are placed with the curved surfaces at a distance d as shown in the figure, with their axes (shown by the dashed line) aligned. When a point source of light P is placed inside rod $S_1$ on its axis at a distance of 50 cm from the curved face, the light rays emanating from it are found to be parallel to the axis inside $S_2$. The distance $d$ is :

• A. 60 cm
• B. 70 cm
• C. 80 cm
• D. 90 cm

Answer 1

Answer: (B)

70 cm

For $S_1$,

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

Substituting the respective values we get,

$\frac{1}{v}-\frac{1.5}{(-50)}=\frac{1-1.5}{-10}$

or

$\frac{1}{v}+\frac{3}{100}=\frac{1}{20}$

$v=50cm$

This acts as image for $S_2$ with distance from it's aperture as $d-50cm$
Given $v=\infty ,u=-(d-50)cm,R=10cm$

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

$0-\frac{1}{-(d-50)}=\frac{1.5-1}{10}$

or

$\frac{1}{d-50}=\frac{1}{20}$

or

$d=70cm$