JEE Main 2026 — Jan 28, Shift 1 — Physics solution

JEE Main2026Jan 28, Shift 1Physics

A convex lens of refractive index 1.5 and focal length $f = 18$ cm is immersed in water. The difference in focal lengths of the given lens when it is in water and in air is $α \times f$ . The value of $α$ is ____. (refractive index of water = 4/3)

Your answer

Solution

Lens-maker's formula: $\frac{1}{f} = (μ_{r e l} - 1) (\frac{1}{R _{1}} + \frac{1}{R _{2}})$ , with $μ_{r e l}$ being the lens-medium relative index.

In air: $\frac{1}{f _{air}} = (1.5 - 1) (\frac{1}{R _{1}} + \frac{1}{R _{2}}) = 0.5 S$ where $S = \frac{1}{R _{1}} + \frac{1}{R _{2}}$ .

In water: $μ_{r e l} = \frac{1.5}{4/3} = \frac{9}{8}$ , so $\frac{1}{f _{w}} = \frac{1}{8} S$ .

Hence $\frac{f _{w}}{f _{air}} = \frac{0.5}{1/8} = 4$ , giving $f_{w} = 4 f_{air} = 4 f$ .

$f_{w} - f_{air} = 4 f - f = 3 f = α f \Rightarrow α = 3$ .

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