JEE Main 2025 Jan 24 Shift 1, Physics Q15: Thin Lens and Mirrors

JEE Main2025Jan 24, Shift 1Physics

A plano-convex lens having radius of curvature of first surface 2 cm exhibits focal length of $f_{1}$ in air. Another plano-convex lens with first surface radius of curvature 3 cm has focal length of $f_{2}$ when it is immersed in a liquid of refractive index 1.2. If both the lenses are made of same glass of refractive index 1.5, the ratio of $f_{1}$ and $f_{2}$ will be :-

A
$3 : 5$
B
$1 : 3$
C
$1 : 2$
D
$2 : 3$

Solution

Lensmaker's formula: $\frac{1}{f} = (μ_{r e l} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$ , with $R_{2} = \infty$ for the plane face.

Lens 1 in air: $μ_{r e l} = 1.5$ , $R_{1} = 2$ cm.

$\frac{1}{f _{1}} = (1.5 - 1) \cdot \frac{1}{2} = \frac{1}{4} \Rightarrow f_{1} = 4$ cm

Lens 2 in liquid of index 1.2: $μ_{r e l} = \frac{1.5}{1.2} = 1.25$ , $R_{1} = 3$ cm.

$\frac{1}{f _{2}} = (1.25 - 1) \cdot \frac{1}{3} = \frac{1}{12} \Rightarrow f_{2} = 12$ cm

$f_{1} : f_{2} = 4 : 12 = 1 : 3$ .

$∴$ Option (2).

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JEE Main 2025 Jan 24 Shift 1, Physics Q15: Thin Lens and Mirrors

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