Two identical glass rods S1 and S2 (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 S1 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 S2. The distance d is

Question

Two identical glass rods S1 and S2 (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 S1 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 S2. The distance d is (A) 60 cm (B) 70 cm (C) 80 cm (D) 90 cm

Tardigrade Admin · Accepted Answer

This question is of multiple event.
First event on curved surface of

S_{1}

& Second on curved surface of

S_{2}

Event

- 1 n_{2} = 1

n_{1} = 1.5

\frac{1}{v} - \frac{1.5}{u} = \frac{1 - 1.5}{R}

\frac{1}{v} - \frac{1.5}{- 50} = \frac{- 0.5}{- 10}

\frac{1}{v} + \frac{1.5}{50} = \frac{1}{20}

\frac{1}{v} = \frac{1}{20} - \frac{1.5}{50}

\frac{1}{v} = \frac{5 - 3}{100} = \frac{2}{100} ∴ v = 50 c m

for second event

u = - (d - 50)

v = \infty

R = + 10 c m

\frac{1.5}{v} - \frac{1}{u} = \frac{1.5 - 1}{R}

\frac{1.5}{\infty} - \frac{1}{- ( d - 50 )} = \frac{0.5}{10}

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

20 = d - 50

d = 70 c m