Question

A slab of thickness 1.5 cm and refractive index 1.5 is introduced between the image and the lens. Find the new position of the object so that image is again formed on the screen

Answer 1

Given: A lens is made of three thin different mediums. Radius of curvature and refractive index of each medium. Surface AB is straight. An object is placed at some distance from the lens by which a real image is formed on the screen placed at a distance of 10 cm from the lens.

To find the new position of the object so that image is again formed on the screen, if a slab of thickness 1.5 cm and refractive index 1.5 is introduced between the image and the lens.

Solution:

Shift in the image away from the slab due to introduction of slab is,

$\Delta x=(1-\frac{1}{n})t=(1-\frac{1}{1.5})\times 1.5=0.5$

The distance of the object so that image is again formed on the screen

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}⟹\frac{1}{9.5}-\frac{1}{u}=\frac{1}{10}⟹u=190cm$