Question

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?

• A. 0.8 m
• B. 3.2 m
• C. 1.2 m
• D. 5.6 m

Answer 1

Answer: (B)

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
$\Delta x = t \left(1-\frac{1}{\mu}\right)=\frac{1}{3} cm$
Now $v =12-\frac{1}{3}=\frac{35}{3} cm$
Again apply lens formula
$\frac{1}{0.12}+\frac{1}{u}=\frac{1}{f}=\frac{210}{24}$
Solving $u =-5.6 m$
Thus shift of object is
$5.6-2.4=3.2 m$