Q68 / 90 · JEE Main 2022 (25 Jul Shift 2) · Physics

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
  • Correct answer
    B
    3.2 m
  • C
    1.2 m
  • D
    5.6 m

Solution

Given: object using the lens formula: \(\begin{aligned}& \frac{1}{f}=\frac{1}{v \prime}-\frac{1}{w^{\prime}} \\\Rightarrow & \frac{21}{240} \mathrm{~cm}=\frac{3}{35} \mathrm{~cm}-\frac{1}{u^{\prime}} \\\Rightarrow \frac{1}{u^{\prime}} & =\frac{3}{35} \mathrm{~cm}-\frac{21}{240} \mathrm{~cm} \\\Rightarrow & \frac{1}{u^{\prime}}=\frac{1}{5}\left[\frac{3}{7} \mathrm{~cm}-\frac{21}{48} \mathrm{~cm}\right] \\\Rightarrow \frac{1}{u^{\prime}} & =\frac{1}{5}\left[\frac{144-147}{336}\right] \\\Rightarrow u^{\prime} & =-560 \mathrm{~cm}=-5.6 \mathrm{~m}\end{aligned}\) Hence, the object should be at \(5.6 \mathrm{~m}\) in front of the lens. Hence, the oblect should be moved by 5.6-24= \(3.2 \mathrm{~cm}\).

Hence, the correct answer is option: 'B'.

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