JEE Main 2023 · PhysicsNumerical answerNumerical valueMediumCalculation

JEE Main 11 April 2023, Shift 1, Physics Q52

Question 52 of 90 in this shift, Physics question 22 of 30, Section B.

The radius of curvature of each surface of a convex lens having refractive index 1.8 is 20 cm. The lens is now immersed in a liquid of refractive index 1.5. The ratio of power of lens in air to its power in the liquid will be

x : 1

. The value of

x

is _________

Official answer

NTA final key.

Chapter: Ray Optics and Optical Instruments
Topic: Lens Maker's Formula
Idea tested: Lens Immersed in a Liquid

Same idea in other shifts

Asked 5× in all

7 Apr 2025, Shift 1 · Q42A lens having refractive index 1.6 has focal length of 12 cm, when it is in air. Find the focal length of the lens when it is placed in…EasySingle correct
8 Apr 2025, Shift 2 · Q41A concave-convex lens of refractive index 1.5 and the radii of curvature of its surfaces are 30 cm and 20 cm, respectively. The concave…HardSingle correct
29 Jan 2024, Shift 1 · Q45A biconvex lens of refractive index 1.5 has a focal length of 20 cm in air. Its focal length when immersed in a liquid of refractive index…MediumSingle correct
16 Mar 2021, Shift 2 · Q3The refractive index of a converging lens is 1.4. What will be the focal length of this lens if it is placed in a medium of same…EasySingle correct

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 42 of the 174 JEE Main shifts held from 2017 to 2026.