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Physics · Transverse waves

JEE Main 2025 — 3 April, Evening Shift — Question 53

Two monochromatic light beams have intensities in the ratio 1:91: 9. An interference pattern is obtained by these beams. The ratio of the intensities of maximum to minimum is

  1. Option A:

    3:13: 1

  2. Option B:

    9:19: 1

  3. Option C:

    8:18: 1

  4. Option D:

    4:14: 1

    Correct

Answer: D

Step-by-step solution

Imax⁡Imin⁡=(I1+I2)2(I1−I2)2=(1+3)2(1−3)2=164=4\frac{I_{\max }}{I_{\min }}=\frac{\left(\sqrt{I_{1}}+\sqrt{I_{2}}\right)^{2}}{\left(\sqrt{I_{1}}-\sqrt{I_{2}}\right)^{2}}=\frac{(1+3)^{2}}{(1-3)^{2}}=\frac{16}{4}=4

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
3 April, Evening Shift
Subject
Physics
Chapter
Transverse waves
Topic
super position, Reflection and Transmission of waves