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MHT CET 2026 (20 May, Shift 1) · Spherical Surfaces & Thin Lenses

Spherical Surfaces & Thin Lenses question from MHT CET 2026 (20 May, Shift 1)

Q11MHT CET 2026 · 20 MAY · SHIFT 1Medium

Spherical Surfaces & Thin Lenses

The focal length of a glass convex lens in air is faf_a and that in water is fwf_w. The refractive index from air to glass is 32\frac{3}{2} and that from air to water is 43\frac{4}{3}. The ratio fa:fwf_a : f_w is

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Answer: (c)

  1. fwfa=μg−1μgμw−1\frac{f_w}{f_a} = \frac{\mu_g - 1}{\frac{\mu_g}{\mu_w} - 1}
  2. μgμw=3/24/3=98\frac{\mu_g}{\mu_w} = \frac{3/2}{4/3} = \frac{9}{8}, so fwfa=1/21/8=4\frac{f_w}{f_a} = \frac{1/2}{1/8} = 4
  3. So the answer is (c) fa:fw=1:4f_a : f_w = 1 : 4.
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