JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Geometrical Optics

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) $5 : 2$ ✓ Correct
(B) $5 : 12$
(C) $12 : 5$
(D) $2 : 5$

Answer: (A)

Solution

Using the lens maker's formula, the power of a lens is:

P = (μ - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})

For the biconvex lens (refractive index 1.5):

P_{A} = 0.5 (\frac{1}{R _{1}} + \frac{1}{R _{2}})

For the plano-concave lens (refractive index 1.7) with one flat surface and radius

R_{2}

P_{B} = 0.7 (\frac{1}{R _{2}})

Given

∣ P_{A} ∣ = ∣ P_{B} ∣

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

\frac{1}{R _{1}} + \frac{1}{R _{2}} = \frac{1.4}{R _{2}}

\frac{1}{R _{1}} = \frac{0.4}{R _{2}}

\frac{R _{1}}{R _{2}} = \frac{1}{0.4} = \frac{5}{2}

JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Single correct option

Practise more: all JEE Main PYQs on Geometrical Optics →

JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Geometrical Optics

Solution

More JEE Main questions on Geometrical Optics