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JEE Physics 2025 Question with Solution

A lens having refractive index 1.61.6 has focal length of 12 cm12 \, \text{cm}, when it is in air. Find the focal length of the lens when it is placed in water. (Take refractive index of water as 1.281.28)

  • A

    355 mm355 \, \text{mm}

  • B

    288 mm288 \, \text{mm}

  • C

    555 mm555 \, \text{mm}

  • D

    655 mm655 \, \text{mm}

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: refractive index of the lens μL=1.6\mu_L = 1.6, focal length in air fair=12 cmf_{\text{air}} = 12 \, \text{cm}, and refractive index of water μm=1.28\mu_m = 1.28.

Find: focal length of the lens in water.

Use the lens maker relation in a medium:

1f=(μLμm−1)(1R1−1R2)\frac{1}{f} = \left(\frac{\mu_L}{\mu_m} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)

For air, μm=1\mu_m = 1, so

112=(1.6−1)(1R1−1R2)\frac{1}{12} = (1.6 - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) 112=610(1R1−1R2)\frac{1}{12} = \frac{6}{10}\left(\frac{1}{R_1} - \frac{1}{R_2}\right) (1R1−1R2)=1072\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = \frac{10}{72}

Now for water:

1f=(1.61.28−1)(1072)\frac{1}{f} = \left(\frac{1.6}{1.28} - 1\right)\left(\frac{10}{72}\right) 1f=14×1072\frac{1}{f} = \frac{1}{4} \times \frac{10}{72} f=28.8 cmf = 28.8 \, \text{cm}

Converting to millimetres,

f=288 mmf = 288 \, \text{mm}

Therefore, the correct option is B.

Using the extracted hint and discrepancy check

Given: the lens is moved from air to water, so the surrounding medium changes.

Find: the new focal length.

The first extracted approach on the page states

fmedium=fair⋅μmediumμlensf_{\text{medium}} = \frac{f_{\text{air}} \cdot \mu_{\text{medium}}}{\mu_{\text{lens}}}

and then obtains 9.6 cm9.6 \, \text{cm} and 96 mm96 \, \text{mm}, but the same approach finally states 288 mm288 \, \text{mm}. This is internally inconsistent.

The second approach uses the correct lens maker relation in a medium and gives

f=28.8 cm=288 mmf = 28.8 \, \text{cm} = 288 \, \text{mm}

So the answer derived from the solution working is 288 mm288 \, \text{mm}, which matches option B.

Common mistakes

  • Using a direct proportionality like fmedium=fairμmediumμlensf_{\text{medium}} = \frac{f_{\text{air}} \mu_{\text{medium}}}{\mu_{\text{lens}}} is incorrect here because the focal length depends on (μLμm−1)\left(\frac{\mu_L}{\mu_m} - 1\right), not only on the ratio of refractive indices. Use the lens maker relation in the medium.

  • Forgetting that the surrounding medium changes the power of the lens is wrong because the curvature term remains the same while the refractive contrast changes. First extract (1R1−1R2)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) from the air case, then use water.

  • Converting 28.8 cm28.8 \, \text{cm} to millimetres incorrectly leads to a wrong option. Since 1 cm=10 mm1 \, \text{cm} = 10 \, \text{mm}, multiply by 1010 to get 288 mm288 \, \text{mm}.

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