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JEE Mains · Physics · STD 12 - Ray optics and optical instruments
  • JEE Mains 2022
  • Hard
  • 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 \(\dots\,m\)

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Verified Solution

Answer & Solution

Correct Answer

(B) \(3.2\)

Step-by-step Solution

Detailed explanation

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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