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JEE Main 2026 (28 Jan, Shift 1) · Spherical Surfaces & Thin Lenses

Spherical Surfaces & Thin Lenses question from JEE Main 2026 (28 Jan, Shift 1)

Q15JEE Main 2026 · 28 JAN · SHIFT 1Medium

Spherical Surfaces & Thin Lenses

A convex lens of refractive index 1.5 and focal length f=18 cmf = 18\text{ cm} is immersed in water. The difference in focal lengths of the given lens when it is in water and in air is α×f\alpha \times f. The value of α\alpha is ______.

(refractive index of water =4/3= 4/3)

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

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Answer: 3

  1. fwf=μ−1μμw−1=0.598−1=4\frac{f_w}{f} = \frac{\mu - 1}{\frac{\mu}{\mu_w} - 1} = \frac{0.5}{\frac{9}{8} - 1} = 4
  2. fw−f=4f−f=3ff_w - f = 4f - f = 3f
  3. So α=3\alpha = 3.
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