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JEE Physics 2025 Question with Solution

In a long glass tube, a mixture of two liquids A and B with refractive indices 1.31.3 and 1.41.4 respectively, forms a convex refractive meniscus towards A. If an object placed at 13 cm13 \, \text{cm} from the vertex of the meniscus in A forms an image with a magnification of −2-2, then the radius of curvature of the meniscus is:

  • A

    13\frac{1}{3} cm

  • B

    11 cm

  • C

    43\frac{4}{3} cm

  • D

    23\frac{2}{3} cm

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: Refractive index of medium AA is μ1=1.3\mu_1 = 1.3, refractive index of medium BB is μ2=1.4\mu_2 = 1.4, object distance is u=−13 cmu = -13 \, \text{cm}, and magnification is m=−2m = -2.

Find: Radius of curvature RR of the convex refracting meniscus.

For refraction at a spherical surface, use the magnification relation:

m=μ1vμ2um = \frac{\mu_1 v}{\mu_2 u}

Substituting the given values:

−2=1.3 v1.4×(−13)-2 = \frac{1.3 \, v}{1.4 \times (-13)}

So,

2=1.3 v18.22 = \frac{1.3 \, v}{18.2} v=28 cmv = 28 \, \text{cm}

Now apply the refraction formula at a spherical surface:

μ2v−μ1u=μ2−μ1R\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}

Substituting the values:

1.428−1.3−13=1.4−1.3R\frac{1.4}{28} - \frac{1.3}{-13} = \frac{1.4 - 1.3}{R} 0.05+0.1=0.1R0.05 + 0.1 = \frac{0.1}{R} 0.15=0.1R0.15 = \frac{0.1}{R} R=0.10.15=23 cmR = \frac{0.1}{0.15} = \frac{2}{3} \, \text{cm}

Therefore, the radius of curvature of the meniscus is 23 cm\frac{2}{3} \, \text{cm}. The correct option is D.

The solution contains an intermediate numerical inconsistency in one approach, but its final conclusion and the stated correct option both give R=23 cmR = \frac{2}{3} \, \text{cm}.

Using sign convention carefully

Given: The surface separates medium AA with μ1=1.3\mu_1 = 1.3 and medium BB with μ2=1.4\mu_2 = 1.4. The object is in medium AA at u=−13 cmu = -13 \, \text{cm}. Magnification is m=−2m = -2.

Find: The radius of curvature RR.

Use the relation for magnification in refraction at a spherical surface:

m=μ1vμ2um = \frac{\mu_1 v}{\mu_2 u}

Hence,

−2=1.3 v1.4×(−13)-2 = \frac{1.3 \, v}{1.4 \times (-13)} v=28 cmv = 28 \, \text{cm}

This shows that the image is formed in medium BB at a positive image distance.

Now substitute in the refraction formula:

μ2v−μ1u=μ2−μ1R\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R} 1.428−1.3−13=0.1R\frac{1.4}{28} - \frac{1.3}{-13} = \frac{0.1}{R} 120+110=0.1R\frac{1}{20} + \frac{1}{10} = \frac{0.1}{R} 320=0.1R\frac{3}{20} = \frac{0.1}{R} R=0.1×203=23 cmR = \frac{0.1 \times 20}{3} = \frac{2}{3} \, \text{cm}

Therefore, the required radius of curvature is 23 cm\frac{2}{3} \, \text{cm}, so the correct option is D.

Common mistakes

  • Using the wrong magnification formula is a common mistake. For refraction at a spherical surface, the refractive indices must be included. Do not use the thin-lens magnification formula directly; use the relation involving μ1\mu_1, μ2\mu_2, uu, and vv.

  • Ignoring sign convention can reverse the result. The object is in medium AA, so u=−13 cmu = -13 \, \text{cm} under the Cartesian sign convention. Treating uu as positive gives an incorrect equation for RR.

  • Confusing 'convex towards A' with the sign of RR without reference to the chosen axis is incorrect. First write the formula with proper sign convention, then substitute the geometry consistently.

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