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 × f. The value of α is
Q. 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 × f. The value of α is ____ .

(refractive index of water = 4 / 3)

Correct Answer: 3

Explanation

For a thin lens, the lens maker formula is:

1 / f = (μ − 1) (1/R₁ − 1/R₂)

In air, refractive index of lens μ = 1.5:

1 / fair = (1.5 − 1)(1/R₁ − 1/R₂)

When the lens is immersed in water, effective refractive index becomes:

μ′ = μlens / μwater = 1.5 ÷ (4/3) = 1.125

So focal length in water:

1 / fwater = (1.125 − 1)(1/R₁ − 1/R₂)

Taking ratio:

fwater / fair = (1.5 − 1) / (1.125 − 1) = 0.5 / 0.125 = 4

Thus:

fwater = 4f

Difference in focal lengths:

fwater − fair = 4f − f = 3f

Hence:

α = 3

Updated for JEE Main 2026: This PYQ is important for JEE Mains, JEE Advanced and other competitive exams. Practice more questions from this chapter.

Scroll to Top