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JEE Main 2019 (9 Apr, Shift 2) · Combination of Lenses

Combination of Lenses question from JEE Main 2019 (9 Apr, Shift 2)

Q56JEE Main 2019 · 9 APR · SHIFT 2Medium

Combination of Lenses

A thin convex lens L (refractive index = 1.5) is placed on a plane mirror M. When a pin is placed at A, such that OA = 18 cm, its real inverted image is formed at A itself, as shown in figure. When a liquid of refractive index μ1 is put between the lens and the mirror, The pin has to be moved to A', such that OA' = 27 cm, to get its inverted real image at A' itself. The value of μ1 will be :-

Figure

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Answer: (d)

  1. Image at the object itself means rays leave the lens parallel and hit the plane mirror normally, so the object is at the focus: f=18 cmf = 18\text{ cm}.
  2. Equiconvex lens: 118=0.5⋅2R⇒R=18 cm\frac{1}{18} = 0.5\cdot\frac{2}{R} \Rightarrow R = 18\text{ cm}
  3. Liquid forms a plano-concave lens: 1fliq=−μ1−118\frac{1}{f_{liq}} = -\frac{\mu_1-1}{18}; new combination focal length is 27 cm.
  4. 127=118−μ1−118⇒μ1−1=13\frac{1}{27} = \frac{1}{18} - \frac{\mu_1-1}{18} \Rightarrow \mu_1 - 1 = \frac{1}{3}
  5. So the answer is (d) μ1=43\mu_1 = \frac{4}{3}.
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