For an object placed at a distance 2.4 m from a lens, a sharp focused image is observed on a screen placed at a distance 12 cm from the lens. A glass plate of refractive index 1.5 and thickness 1 cm is introduced between lens and screen such that the glass plate plane faces parallel to the screen. By what distance should the object be shifted so that a sharp focused image is observed again on the screen?

see full answer

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

1.2 m

5.6 m

0.8 m

3.2 m

answer is B.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Applying lens formula

$\frac{1}{0 .12} + \frac{1}{2 .4} = \frac{1}{f} \Rightarrow \frac{1}{f} = \frac{210}{24}$

Upon putting the glass slab, shift of image is

$Δx = t (1 - \frac{1}{μ}) = \frac{1}{3} cm Now v = 12 - \frac{1}{3} = \frac{35}{3} cm$

Again apply lens formula

$\frac{3}{0.35} - \frac{1}{u} = \frac{1}{f} = \frac{210}{24}$

Solving $u = - 5.6 m$

Thus shift of object is

$5 .6 - 2 .4 = 3 .2 m$

Watch 3-min video & get full concept clarity