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MCQMediumJEE 2023Refraction & Lenses

JEE Physics 2023 Question with Solution

The radius of curvature of each surface of a convex lens having refractive index 1.81.8 is 20 cm20 \, \text{cm}. The lens is now immersed in a liquid of refractive index 1.51.5. The ratio of power of lens in air to its power in the liquid will be:

  • A

    44

  • B

    33

  • C

    1.51.5

  • D

    22

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: Refractive index of lens is 1.81.8, refractive index of liquid is 1.51.5, and radius of curvature of each surface is 20 cm20 \, \text{cm}.

Find: The ratio of power of lens in air to its power in liquid.

Use the lens maker relation in a medium:

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

For a symmetric convex lens,

R1=+R,R2=−RR_1 = +R, \qquad R_2 = -R

So,

(1R1−1R2)=1R−(−1R)=2R\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = \frac{1}{R} - \left(-\frac{1}{R}\right) = \frac{2}{R}

In air,

Pair=(1.8−1)2R=0.82RP_{\text{air}} = (1.8 - 1)\frac{2}{R} = 0.8\frac{2}{R}

In liquid,

Pliquid=(1.81.5−1)2R=(1.2−1)2R=0.22RP_{\text{liquid}} = \left(\frac{1.8}{1.5} - 1\right)\frac{2}{R} = (1.2 - 1)\frac{2}{R} = 0.2\frac{2}{R}

Therefore,

PairPliquid=0.82R0.22R=4\frac{P_{\text{air}}}{P_{\text{liquid}}} = \frac{0.8\frac{2}{R}}{0.2\frac{2}{R}} = 4

Therefore, the ratio of power in air to power in liquid is 44. The correct option is A.

The solution shows intermediate working leading to 22, but that working uses an incorrect medium formula. Using the correct lens maker formula in a medium gives 44, which matches the stated correct answer.

Common mistakes

  • Using nlens−nmediumn_{\text{lens}} - n_{\text{medium}} instead of nlensnmedium−1\frac{n_{\text{lens}}}{n_{\text{medium}}} - 1 for a lens in a medium is incorrect. The refractive index must be taken relative to the surrounding medium. Use the relative refractive index form in the lens maker formula.

  • Taking both radii with the same sign is wrong for a convex lens. For the standard sign convention, use R1=+RR_1 = +R and R2=−RR_2 = -R, so that 1R1−1R2=2R\frac{1}{R_1} - \frac{1}{R_2} = \frac{2}{R}.

  • Comparing focal lengths directly without tracking that power is P=1fP = \frac{1}{f} can cause inversion errors. First find power in each medium, then take the ratio PairPliquid\frac{P_{\text{air}}}{P_{\text{liquid}}}.

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