JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Geometrical Optics

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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
(B)3.2 m✓ Correct
(C)1.2 m
(D)5.6 m

Answer: (B)

Solution

Initially, using the lens formula with

u = - 2.4

m and

v = 0.12

\frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{1}{0.12} + \frac{1}{- 2.4} = \frac{210}{24}

When the glass plate is introduced, the image shifts by:

Δ x = t (1 - \frac{1}{μ}) = 1 (1 - \frac{1}{1.5}) = \frac{1}{3} cm

New image distance:

v^{'} = 12 - \frac{1}{3} = \frac{35}{3}

cm

Using lens formula again:

\frac{1}{f} = \frac{1}{v ^{'}} + \frac{1}{u ^{'}} ⟹ \frac{210}{24} = \frac{3}{35} + \frac{1}{u ^{'}}

Solving gives

u^{'} = - 5.6

m.

Shift in object position:

∣ u^{'} ∣ - ∣ u ∣ = 5.6 - 2.4 = 3.2

JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Single correct option

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JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Geometrical Optics

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