Physics · Geometrical Optics

JEE Main 2026 — 28 January, Evening Shift — Question 28

A biconvex lens is formed by using two thin planoconvex lenses, as shown in the figure. The refractive index and radius of curved surfaces are also mentioned in figure. When an object is placed on the left side of lens at a distance of 30 cm from the biconvex lens, the magnification of the image will be :

Option A:
$-2$
Correct
Option B:
$+2$
Option C:
$+2.5$
Option D:
$-2.5$

Answer: A

Step-by-step solution

$\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}_{\mathrm{net}}}=\frac{1}{\mathrm{f}_{1}}+\frac{1}{\mathrm{f}_{2}}$ $\frac{1}{\mathrm{v}}+\frac{1}{30}=(1.5-1)\left(\frac{1}{15}-\frac{1}{\infty}\right)+(1.2-1)\left(\frac{1}{\infty}+\frac{1}{12}\right)$ $\frac{1}{\mathrm{v}}+\frac{1}{30}=\frac{1}{30}+\frac{1}{60}$ $\mathrm{v}=60$ $\mathrm{m}=\frac{\mathrm{v}}{\mathrm{u}}=\frac{60}{-30}=-2$

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

For a transparent prism, if the angle of minimum deviation is equal to its refracting angle, the refractive index n of the prism satisfies.

Answer: A

Step-by-step solution

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