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MHT CET 2023 (13 May, Shift 1) · Spherical Surfaces & Thin Lenses

Spherical Surfaces & Thin Lenses question from MHT CET 2023 (13 May, Shift 1)

Q54MHT CET 2023 · 13 MAY · SHIFT 1Medium

Spherical Surfaces & Thin Lenses

A spherical surface of radius of curvature 'RR' separates air from glass of refractive index 1.5. The centre of curvature is in the glass. A point object PP placed in air forms a real image QQ in the glass. The line PQPQ cuts the surface at point 'OO' and PO=OQ=xPO = OQ = x. Hence the distance 'xx' is equal to

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Answer: (d)

  1. Refraction at a spherical surface: μ2v−μ1u=μ2−μ1R\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}
  2. u=−xu = -x, v=+xv = +x, R=+RR = +R: 1.5x+1x=0.5R\frac{1.5}{x} + \frac{1}{x} = \frac{0.5}{R}
  3. 2.5x=0.5R⇒x=5R\frac{2.5}{x} = \frac{0.5}{R} \Rightarrow x = 5R
  4. So the answer is (d) 5 R5\,R.
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