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Physics · Geometrical Optics

JEE Main 2026 — 28 January, Morning Shift — Question 42

A convex lens of refractive index 1.5 and focal length f=18 cmf=18 \mathrm{~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\alpha \times \mathrm{f}. The value of α\alpha is ____\_\_\_\_ . (refractive index of water =4/3=4 / 3 )

Answer: 3

Numerical answer — enter this value.

Step-by-step solution

1fAir =(1.5−11)(1R1+1R2)\frac{1}{\mathrm{f}_{\text {Air }}}=\left(\frac{1.5-1}{1}\right)\left(\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}\right) 1fwater =(1.5−4/34/3)(1R1+1R2)\frac{1}{\mathrm{f}_{\text {water }}}=\left(\frac{1.5-4 / 3}{4 / 3}\right)\left(\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}\right) ⇒fwater fair =0.50.5/4=4\Rightarrow \frac{\mathrm{f}_{\text {water }}}{\mathrm{f}_{\text {air }}}=\frac{0.5}{0.5 / 4}=4 ⇒fwater −fair =3f\Rightarrow \mathrm{f}_{\text {water }}-\mathrm{f}_{\text {air }}=3 \mathrm{f}

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
28 January, Morning Shift
Subject
Physics
Chapter
Geometrical Optics
Topic
Lenses and Their Combinations, Silvering of Lens

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