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Physics · Geometrical Optics

JEE Main 2025 — 28 January, Evening Shift — Question 52

In a long glass tube, mixture of two liquids A and B with refractive indices 1.3 and 1.4 respectively, forms a convex refractive meniscus towards A. If an object placed at 13 cm from the vertex of the meniscus in A forms an image with a magnification of '–2' then the radius of curvature of meniscus is :

  1. Option A:

    1 cm

  2. Option B:

    1/3 cm

  3. Option C:

    2/3 cm

    Correct
  4. Option D:

    4/3 cm

Answer: C

Step-by-step solution

n2v−n1v−n2−n1R,,1.4v−1.3−13=0.1R\frac{{{n}_{2}}}{v}-\frac{{{n}_{1}}}{v}-\frac{{{n}_{2}}-{{n}_{1}}}{R},,\frac{1.4}{v}-\frac{1.3}{-13}=\frac{0.1}{R} 1.4v=1−R10R\frac{1.4}{v}=\frac{1-R}{10R} m =v/n2 u/n1  -2×( -13)1.3= 10R1−R m\text{ =}\frac{\text{v/}{{n}_{2}}}{\text{ u/}{{n}_{1}}\text{ }}\text{ -2}\times \frac{\left( \text{ -}13 \right)}{1.3}\text{= }\frac{10R}{1-R}\text{ } R=23cmR=\frac{2}{3}cm
Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
28 January, Evening Shift
Subject
Physics
Chapter
Geometrical Optics
Topic
Refraction at Curved Surface and Glass Sphere

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