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JEE Main 2023 (8 Apr, Shift 2) · Spherical Surfaces & Thin Lenses

Spherical Surfaces & Thin Lenses question from JEE Main 2023 (8 Apr, Shift 2)

Q55JEE Main 2023 · 8 APR · SHIFT 2Medium

Spherical Surfaces & Thin Lenses

Two transparent media having refractive indices 1.0 and 1.5 are separated by a spherical refracting surface of radius of curvature 30 cm30\text{ cm}. The centre of curvature of surface is towards denser medium and a point object is placed on the principal axis in rarer medium at a distance of 15 cm15\text{ cm} from the pole of the surface. The distance of image from the pole of the surface is ____________ cm\text{cm}.

Numerical answer type. Work it out, then open the answer.

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Answer: 30

  1. u=−15u = -15 cm, R=+30R = +30 cm (centre in denser medium), μ1=1\mu_1 = 1, μ2=1.5\mu_2 = 1.5.
  2. 1.5v+115=0.530=160\frac{1.5}{v} + \frac{1}{15} = \frac{0.5}{30} = \frac{1}{60}
  3. 1.5v=160−460=−120⇒v=−30\frac{1.5}{v} = \frac{1}{60} - \frac{4}{60} = -\frac{1}{20} \Rightarrow v = -30 cm (virtual, in rarer medium).
  4. So the image distance is 30 cm.
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