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

A convex lens of refractive index 1.5 and focal length

f = 18 cm

is immersed in water. The difference in focal lengths of the given lens when it is in water and in air is

α \times f

. The value of

α

is ____.
(refractive index of water

= 4/3

)

Answer: 3

Solution

Using the lens maker's formula:

In air:

\frac{1}{f _{ai r}} = (n - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}}) = (1.5 - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})

In water:

\frac{1}{f _{w a t er}} = (\frac{n _{l e n s}}{n _{w a t er}} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}}) = (\frac{1.5}{4/3} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})

Taking the ratio:

\frac{f _{w a t er}}{f _{ai r}} = \frac{0.5}{0.5/4} = 4

Therefore:

f_{w a t er} = 4 f_{ai r} = 4 \times 18 = 72 cm

Difference:

f_{w a t er} - f_{ai r} = 72 - 18 = 54 = 3 \times 18 = 3 f

Hence:

α = 3

JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Integer answer