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JEE Main 2022 (25 Jul, Shift 2) · Refraction, Glass Slab & Apparent Depth

Refraction, Glass Slab & Apparent Depth question from JEE Main 2022 (25 Jul, Shift 2)

Q53JEE Main 2022 · 25 JUL · SHIFT 2Difficult

Refraction, Glass Slab & Apparent Depth

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?

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Answer: (b)

  1. Lens: 1f=1v−1u=112+1240=21240⇒f=24021 cm\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{12} + \frac{1}{240} = \frac{21}{240} \Rightarrow f = \frac{240}{21}\text{ cm}.
  2. The plate shifts the image away by t(1−1μ)=1(1−11.5)=13 cmt\left(1 - \frac{1}{\mu}\right) = 1\left(1 - \frac{1}{1.5}\right) = \frac{1}{3}\text{ cm}.
  3. For a sharp image on the screen, the lens must form its image at v′=12−13=353 cmv' = 12 - \frac{1}{3} = \frac{35}{3}\text{ cm}.
  4. 1u′=335−21240=−1560⇒u′=−560 cm=−5.6 m\frac{1}{u'} = \frac{3}{35} - \frac{21}{240} = -\frac{1}{560} \Rightarrow u' = -560\text{ cm} = -5.6\text{ m}
  5. Shift =5.6−2.4=3.2 m= 5.6 - 2.4 = 3.2\text{ m} (away from the lens).
  6. So the answer is (b) 3.2 m3.2\text{ m}.
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