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JEE Physics 2025 Question with Solution

What is the relative decrease in focal length of a lens for an increase in optical power by 0.1 D0.1 \, \text{D} from 2.5 D2.5 \, \text{D}? ('D\text{D}' stands for dioptre).

  • A

    0.040.04

  • B

    0.400.40

  • C

    0.10.1

  • D

    0.010.01

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: Initial optical power is P1=2.5 DP_1 = 2.5 \, \text{D} and the increase in power is 0.1 D0.1 \, \text{D}.

Find: The relative decrease in focal length.

The optical power PP of a lens is given by

P=1fP = \frac{1}{f}

where PP is the power in dioptres and ff is the focal length in metres.

Initially,

f1=1P1=12.5=0.4 mf_1 = \frac{1}{P_1} = \frac{1}{2.5} = 0.4 \, \text{m}

After the increase in power,

P2=2.6 DP_2 = 2.6 \, \text{D}

so the new focal length is

f2=1P2=12.6≈0.3846 mf_2 = \frac{1}{P_2} = \frac{1}{2.6} \approx 0.3846 \, \text{m}

The decrease in focal length is

Δf=f1−f2=0.4−0.3846=0.0154 m\Delta f = f_1 - f_2 = 0.4 - 0.3846 = 0.0154 \, \text{m}

Therefore, the relative decrease in focal length is

Δff1=0.01540.4=0.0385\frac{\Delta f}{f_1} = \frac{0.0154}{0.4} = 0.0385

Rounding to two decimal places gives 0.040.04. Therefore, the correct option is A.

Step-by-step Calculation

Given: P1=2.5 DP_1 = 2.5 \, \text{D} and the power increases by 0.1 D0.1 \, \text{D}.

Find: Relative decrease in focal length.

Step 1: Calculate the initial focal length using

P=1fP = \frac{1}{f}

Thus,

f1=1P1=12.5=0.4 mf_1 = \frac{1}{P_1} = \frac{1}{2.5} = 0.4 \, \text{m}

Step 2: The new optical power becomes

P2=2.5+0.1=2.6 DP_2 = 2.5 + 0.1 = 2.6 \, \text{D}

Hence,

f2=1P2=12.6≈0.3846 mf_2 = \frac{1}{P_2} = \frac{1}{2.6} \approx 0.3846 \, \text{m}

Step 3: Compute the relative decrease:

Relative decrease=f1−f2f1=0.4−0.38460.4=0.01540.4≈0.04\text{Relative decrease} = \frac{f_1 - f_2}{f_1} = \frac{0.4 - 0.3846}{0.4} = \frac{0.0154}{0.4} \approx 0.04

Therefore, the correct answer is 0.040.04, which corresponds to option A.

Common mistakes

  • Using the increase in power 0.10.1 directly as the relative decrease in focal length is incorrect because focal length is inversely related to power through P=1fP = \frac{1}{f}. First calculate the two focal lengths, then compare them.

  • Computing relative decrease as f1−f2f2\frac{f_1-f_2}{f_2} is wrong because relative decrease must be measured with respect to the initial value. Use f1−f2f1\frac{f_1-f_2}{f_1} instead.

  • Forgetting to update the final power to 2.6 D2.6 \, \text{D} leads to no change in focal length. Always add the increase in power to the initial optical power before calculating the new focal length.

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