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JEE Main 2025 (29 Jan, Shift 2) · Spherical Surfaces & Thin Lenses

Spherical Surfaces & Thin Lenses question from JEE Main 2025 (29 Jan, Shift 2)

Q21JEE Main 2025 · 29 JAN · SHIFT 2Easy

Spherical Surfaces & Thin Lenses

A convex lens made of glass (refractive index = 1.5) has focal length 24 cm in air. When it is totally immersed in water (refractive index = 1.33), its focal length changes to

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Answer: (a)

  1. fwfa=μ−1μμw−1\frac{f_w}{f_a} = \frac{\mu - 1}{\frac{\mu}{\mu_w} - 1}, taking μw=1.33≈43\mu_w = 1.33 \approx \frac{4}{3}
  2. μμw=1.54/3=98\frac{\mu}{\mu_w} = \frac{1.5}{4/3} = \frac{9}{8}, so fw24=0.50.125=4\frac{f_w}{24} = \frac{0.5}{0.125} = 4
  3. fw=96 cmf_w = 96\text{ cm}
  4. So the answer is (a) 96 cm.
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