An object 2.4 m in front of a lens forms a sharp image on a film 12… | JEE Main 2012 (AIEEE)

Q73JEE Main 2012 · AIEEEMedium

Refraction, Glass Slab & Apparent Depth

An object $2.4\text{ m}$ in front of a lens forms a sharp image on a film $12\text{ cm}$ behind the lens. A glass plate $1\text{ cm}$ thick, of refractive index $1.50$ is interposed between lens and film with its plane faces parallel to film. At what distance (from lens) should object shifted to be in sharp focus of film?

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Answer: (d)

Lens: $\dfrac{1}{f} = \dfrac{1}{12} + \dfrac{1}{240} = \dfrac{21}{240}$ (cm).
Plate shifts the image away by $1\left(1 - \dfrac{1}{1.5}\right) = \dfrac{1}{3}$ cm, so the lens alone must image at $v = 12 - \dfrac{1}{3} = \dfrac{35}{3}$ cm.
$\dfrac{1}{u} = \dfrac{21}{240} - \dfrac{3}{35} = \dfrac{735 - 720}{8400} = \dfrac{1}{560}$ .
$u = 560$ cm $= 5.6$ m, so the answer is (d).

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Refraction, Glass Slab & Apparent Depth question from JEE Main 2012 (AIEEE)

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