Physics · Geometrical Optics

JEE Main 2024 — 29 January, Shift 1 — Question 43

A 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 1.6 will be

Option A:
-16 cm
Option B:
-160 cm
Correct
Option C:
+160 cm
Option D:
$+16 \mathrm{~cm}$

Answer: B

Step-by-step solution

$\mu_{1}=1.5$

$\mu_{\mathrm{m}}=1.6$

$\mathrm{f}_{\mathrm{a}}=20 \mathrm{~cm}$

As $\frac{f_{m}}{f_{a}}=\frac{\left(\mu_{1}-1\right) \mu_{m}}{\left(\mu_{1}-\mu_{m}\right)}$ $\frac{\mathrm{f}_{\mathrm{m}}}{20}=\frac{(1.5-1) 1.6}{(1.5-1.6)}$

$\mathrm{f}_{\mathrm{m}}=-160 \mathrm{~cm}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 29 January, Shift 1
Subject: Physics
Chapter: Geometrical Optics
Topic: Lenses and Their Combinations, Silvering of Lens

A convex mirror of radius of curvature 30 cm forms an image that is half the size of the object. The object distance is :

Answer: B

Step-by-step solution

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