JEE Advanced · 2022 · 28 Aug 2022 · Paper 1 · Geometrical Optics

A rod of length

2 c m

makes an angle

\frac{2 π}{3}

rad with the principal axis of a thin convex lens. The lens has a focal length of

10 c m

and is placed at a distance of

\frac{40}{3} c m

from the object as shown in the figure. The height of the image is

\frac{30 3}{13} c m

and the angle made by it with respect to the principal axis is

α

rad. The value of

α

\frac{π}{n} r a d

, where

n

is ____.

Answer: 6

Solution

For the vertical component of the rod at distance

u = - 40/3

cm:

Using lens formula:

\frac{1}{v} - \frac{1}{u} = \frac{1}{f}

\frac{1}{v} = \frac{1}{10} - \frac{3}{40} = \frac{1}{40}

v = 40

cm

For the end of the rod:

\frac{h _{i}}{h _{0}} = \frac{v}{u} \Rightarrow \frac{- 30 3 /13}{3} = \frac{v _{1}}{- 43/3}

Giving

v_{1} = 430/13

cm

Horizontal displacement:

x = 40 - 430/13 = 90/13

cm

Angle:

tan α = \frac{30 3 /13}{90/13} = \frac{1}{3}

Therefore

α = 30° = \frac{π}{6}

, so

n = 6

JEE Advanced · 2022 · 28 Aug 2022 · Paper 1 · Numerical answer