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JEE Main 202225 JulyEvening Shift
Exam
JEE Main
Year
2022
Date
25 July
Shift
Evening Shift
Subject
Physics
Chapter
Ray Optics

JEE Main 2022 · 25 July · Evening Shift · Physics · Ray Optics

For an object placed at a distance 2 . 4m from a lens, a sharp...

For an object placed at a distance 2.4m2 . 4m from a lens, a sharp focused image is observed on a screen placed at a distance 12cm12cm from the lens. A glass plate of refractive index 1.51 . 5 and thickness 1cm1cm 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?

Options

  1. A0.8m0 . 8m
  2. B3.2m3 . 2m
  3. C1.2m1 . 2m
  4. D5.6m5 . 6m

Correct answer

B. 3.2m3 . 2m

Step-by-step solution

Given: object using the lens formula: (1f=1v′−1w′⇒21240 cm=335 cm−1u′⇒1u′=335 cm−21240 cm⇒1u′=15[37 cm−2148 cm]⇒1u′=15[144−147336]⇒u′=−560 cm=−5.6 m\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}).
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