MCQ
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 $\dots\,m$
  • A
    $0.8$
  • $3.2$
  • C
    $1.2$
  • D
    $5.6$

Answer

Correct option: B.
$3.2$
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

$5.6-2.4=3.2\,m$

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