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

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

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

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Answer: 3

$\frac{f_w}{f} = \frac{\mu - 1}{\frac{\mu}{\mu_w} - 1} = \frac{0.5}{\frac{9}{8} - 1} = 4$
$f_w - f = 4f - f = 3f$
So $\alpha = 3$ .

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