Logo
Jeeify
MCQEasyJEE 2024Refraction & Lenses

JEE Physics 2024 Question with Solution

A thin lens made of material of refractive index 1.51.5 has focal length ff in air. Its focal length in a medium of refractive index 1.31.3 will be:

  • A

    f×0.5f \times 0.5

  • B

    f×1.5f \times 1.5

  • C

    f×1.3f \times 1.3

  • D

    f×2.5f \times 2.5

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: Refractive index of the lens is μl=1.5\mu_l = 1.5. Refractive index of the medium in the question is μm=1.3\mu_m = 1.3. Focal length in air is ff.

Find: The focal length of the lens in the medium.

From the solution, the governing relation used is

fmfa=μl−1μl−μm\frac{f_m}{f_a} = \frac{\mu_l - 1}{\mu_l - \mu_m}

For this question, fa=ff_a = f.

Substitute the values from the question:

fmf=1.5−11.5−1.3\frac{f_m}{f} = \frac{1.5 - 1}{1.5 - 1.3} fmf=0.50.2=2.5\frac{f_m}{f} = \frac{0.5}{0.2} = 2.5

Hence,

fm=2.5ff_m = 2.5f

Therefore, the focal length in the medium is f×2.5f \times 2.5.

Discrepancy note: the solution text is for a different medium value in parts of the page and incorrectly labels option A. Using the same formula shown there with the actual question data μm=1.3\mu_m = 1.3 gives fm=2.5ff_m = 2.5f, so the defensible correct option is D.

Common mistakes

  • Using the medium refractive index from the mismatched the solution instead of the question. That is wrong because the question clearly gives 1.31.3, not 1.61.6. Always substitute values from the question being solved.

  • Applying the lens maker relation with the wrong denominator, such as using μm−μl\mu_m - \mu_l instead of μl−μm\mu_l - \mu_m in the ratio form used here. This changes the sign and magnitude. Write the formula carefully before substitution.

  • Treating the new focal length as directly proportional only to the medium refractive index. That is wrong because the dependence comes through the relative refractive index of lens and medium. Use the full relation, not a visual guess from 1.31.3 and 1.51.5.

Practice more Refraction & Lenses questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Refraction & Lenses

Related questions