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JEE Physics 2024 Question with Solution

A biconvex lens of refractive index 1.51.5 has a focal length of 20 cm20 \, \text{cm} in air. Its focal length when immersed in a liquid of refractive index 1.61.6 will be:

  • A

    −16 cm-16 \, \text{cm}

  • B

    −160 cm-160 \, \text{cm}

  • C

    +160 cm+160 \, \text{cm}

  • D

    +16 cm+16 \, \text{cm}

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: Refractive index of lens is nlens=1.5n_{\text{lens}} = 1.5, focal length in air is fair=20 cmf_{\text{air}} = 20 \, \text{cm}, and refractive index of liquid is nmedium=1.6n_{\text{medium}} = 1.6.

Find: Focal length of the lens in the liquid.

Use the lens maker's formula:

1f=(nlensnmedium−1)(1R1−1R2)\frac{1}{f} = \left( \frac{n_{\text{lens}}}{n_{\text{medium}}} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)

For air:

120=(1.51−1)(1R1−1R2)\frac{1}{20} = \left( \frac{1.5}{1} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) 120=0.5(1R1−1R2)\frac{1}{20} = 0.5 \left( \frac{1}{R_1} - \frac{1}{R_2} \right) (1R1−1R2)=110 cm\left( \frac{1}{R_1} - \frac{1}{R_2} \right) = \frac{1}{10 \, \text{cm}}

Now in liquid:

1fliquid=(1.51.6−1)(110)\frac{1}{f_{\text{liquid}}} = \left( \frac{1.5}{1.6} - 1 \right) \left( \frac{1}{10} \right) 1fliquid=(−116)(110)=−1160\frac{1}{f_{\text{liquid}}} = \left( -\frac{1}{16} \right) \left( \frac{1}{10} \right) = -\frac{1}{160} fliquid=−160 cmf_{\text{liquid}} = -160 \, \text{cm}

Therefore, the focal length of the lens in the liquid is −160 cm-160 \, \text{cm}.

The solution working gives −160 cm-160 \, \text{cm}, which matches option B. Although one the solution states "The Correct Option is A", that conflicts with the shown calculation and the listed options. Hence the correct option is B.

Ratio Method

Given: μl=1.5\mu_l = 1.5, μm=1.6\mu_m = 1.6, and fa=20 cmf_a = 20 \, \text{cm}.

Find: Focal length in the medium.

Use the relation shown in the solution:

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

Substitute the values:

fm20=1.5−11.5−1.6\frac{f_m}{20} = \frac{1.5 - 1}{1.5 - 1.6} fm20=0.5−0.1=−5\frac{f_m}{20} = \frac{0.5}{-0.1} = -5 fm=20×(−5)=−160 cmf_m = 20 \times (-5) = -160 \, \text{cm}

Therefore, the correct option is B.

Common mistakes

  • Using the refractive index of the liquid as if it only changes the magnitude and not the sign. Here nmedium>nlensn_{\text{medium}} > n_{\text{lens}}, so the factor becomes negative and the focal length changes sign. Always check whether the lens remains converging or becomes diverging in the new medium.

  • Substituting directly into the lens maker's formula without first extracting (1R1−1R2)\left( \frac{1}{R_1} - \frac{1}{R_2} \right) from the air case. The curvatures do not change when the lens is immersed; only the refractive-index factor changes.

  • Trusting the displayed option label in the solution instead of the actual calculation. The heading says option A, but the worked result is −160 cm-160 \, \text{cm}, which corresponds to option B. Always verify with the numerical result.

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