Physics · Geometrical Optics

JEE Main 2024 — 6 April, Shift 2 — Question 40

For the thin convex lens, the radii of curvature are at 15 cm and 30 cm respectively. The focal length the lens is 20 cm . The refractive index of the material is :

Option A:
1.2
Option B:
1.4
Option C:
1.5
Correct
Option D:
1.8

Answer: C

Step-by-step solution

$\quad \frac{1}{\mathrm{f}}=\left(\frac{\mu_{\text {lens }}}{\mu_{\text {air }}}-1\right)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)$

$\Rightarrow \frac{1}{+20}=\left(\frac{\mu}{1}-1\right)\left(\frac{1}{+15}-\frac{1}{(-30)}\right)$

$\Rightarrow \frac{1}{20}=(\mu-1)\left(\frac{3}{30}\right)$ $\Rightarrow \mu-1=\frac{1}{2}$

$\Rightarrow \mu=1+\frac{1}{2}=\frac{3}{2}=1 \cdot 5$

Answer key and solution verified before publishing.

Practise Geometrical Optics

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2024
Paper: 6 April, Shift 2
Subject: Physics
Chapter: Geometrical Optics
Topic: Lenses and Their Combinations, Silvering of Lens