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

JEE Main2026Jan 28, Shift 1Physics

The magnitudes of power of a biconvex lens (refractive index 1.5) and that of a plano-concave lens (refractive index = 1.7) are same. If the curvature of plano-concave lens exactly matches with the curvature of back surface of the biconvex lens, then ratio of radius of curvature of front and back surface of the biconvex lens is ______ .

A
5 : 2
B
5 : 12
C
12 : 5
D
2 : 5

Solution

Use the lens-maker's formula. For the biconvex lens (with $R_{1}$ for the front surface, $R_{2}$ for the back), $∣ P_{A} ∣ = (1.5 - 1) (\frac{1}{R _{1}} + \frac{1}{R _{2}}) = 0.5 (\frac{1}{R _{1}} + \frac{1}{R _{2}})$ .

For the plano-concave lens (one curved surface matching $R_{2}$ , the other flat), $∣ P_{B} ∣ = (1.7 - 1) \frac{1}{R _{2}} = \frac{0.7}{R _{2}}$ .

Setting $∣ P_{A} ∣ = ∣ P_{B} ∣$ :

$0.5 (\frac{1}{R _{1}} + \frac{1}{R _{2}}) = \frac{0.7}{R _{2}}$

$\frac{0.5}{R _{1}} = \frac{0.7 - 0.5}{R _{2}} = \frac{0.2}{R _{2}}$

$\frac{R _{2}}{R _{1}} = \frac{0.2}{0.5} = \frac{2}{5} \Rightarrow \frac{R _{1}}{R _{2}} = \frac{5}{2}$ .

Option (1).

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