A plano-convex lens having radius of curvature of first surface 2 cm exhibits focal length of $\mathrm{f}_{1}$ in air. Another plano-convex lens with first surface radius of curvature 3 cm has focal length of $\mathrm{f}_{2}$ 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 $\mathrm{f}_{1}$ and $\mathrm{f}_{2}$ will be :-

JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Geometrical Optics

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 :-

Answer: (B)

For a plano-convex lens in a surrounding medium of index

n_{m}

, with glass index

n_{g} = 1.5

and one flat surface (

R_{2} \to \infty

), the lens maker's formula gives

\frac{1}{f} = (\frac{n _{g}}{n _{m}} - 1) \frac{1}{R}

For the first lens, in air (

n_{m} = 1

R = 2

cm:

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

cm

For the second lens, immersed in a liquid of index 1.2,

R = 3

cm:

\frac{1}{f _{2}} = (\frac{1.5}{1.2} - 1) (\frac{1}{3}) = \frac{0.3}{1.2} \times \frac{1}{3} = \frac{1}{12} \Rightarrow f_{2} = 12

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

JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Single correct option