JEE Main 2025 Apr 2 Shift 2, Physics Q17: Thin Lens and Mirrors

JEE Main2025Apr 2, Shift 2Physics

A bi-convex lens has radius of curvature of both the surfaces same as $1/6$ cm. If this lens is required to be replaced by another convex lens having different radii of curvatures on both sides ( $R_{1} \neq = R_{2}$ ), without any change in lens power then possible combination of $R_{1}$ and $R_{2}$ is :

A
$\frac{1}{3}$ cm and $\frac{1}{3}$ cm
B
$\frac{1}{5}$ cm and $\frac{1}{7}$ cm
C
$\frac{1}{3}$ cm and $\frac{1}{7}$ cm
D
$\frac{1}{6}$ cm and $\frac{1}{9}$ cm

Solution

Using the Lensmaker's equation with the sign convention for a bi-convex lens ( $R_{1} > 0$ , $R_{2} < 0$ ):

$\frac{1}{f} = (μ - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}}) = (μ - 1) (\frac{1}{R _{1}} + \frac{1}{∣ R _{2} ∣})$ .

For the original lens with both radii equal to $R = 1/6$ cm:

$\frac{1}{f} = (μ - 1) (\frac{2}{R}) = (μ - 1) \cdot 12 cm^{- 1}$ .

For an equal-power replacement we need $\frac{1}{R _{1}} + \frac{1}{∣ R _{2} ∣} = \frac{2}{R} = 12 cm^{- 1}$ .

Testing $R_{1} = 1/5$ cm and $∣ R_{2} ∣ = 1/7$ cm: $5 + 7 = 12$ . ✓

The other options either give $6$ , $10$ or $15$ instead of $12$ , so option (2) is the correct match.

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JEE Main 2025 Apr 2 Shift 2, Physics Q17: Thin Lens and Mirrors

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