JEE Advanced · 2015 · 24 May 2015 · Paper 1 · Geometrical Optics

Two idenctical 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 :

Answer: (B)

Solution

For refraction at the curved surface of rod

S_{2}

where rays become parallel inside:

\frac{n}{\infty} - \frac{1}{u _{2}} = \frac{n - 1}{R}

With

n = 1.5

and

R = + 10

cm:

- \frac{1}{u _{2}} = \frac{0.5}{10} = 0.05

u_{2} = - 20

cm

This means the image formed by

S_{1}

must be at 20 cm to the left of the curved surface of

S_{2}

, so

v_{1} = d - 20

.

For refraction at the curved surface of rod

S_{1}

\frac{1}{v _{1}} - \frac{n}{u _{1}} = \frac{1 - n}{R}

With

u_{1} = - 50

cm,

n = 1.5

R = - 10

cm:

\frac{1}{d - 20} + \frac{1.5}{50} = \frac{- 0.5}{- 10} = 0.05

\frac{1}{d - 20} = 0.05 - 0.03 = 0.02

d - 20 = 50

d = 70

JEE Advanced · 2015 · 24 May 2015 · Paper 1 · One or more correct options