Physics Circle home Student Login

Home›Chapters›Wave Optics PYQs›Superposition, Coherence & Intensity›Question 24

JEE Main 2025 (3 Apr, Shift 1) · Superposition, Coherence & Intensity

Superposition, Coherence & Intensity question from JEE Main 2025 (3 Apr, Shift 1)

Q24JEE Main 2025 · 3 APR · SHIFT 1Easy

Superposition, Coherence & Intensity

Two coherent monochromatic light beams of intensities 4I and 9I are superimposed. The difference between the maximum and minimum intensities in the resulting interference pattern is xIxI. The value of xx is___________.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 24

  1. Imax=(4I+9I)2=(2+3)2I=25II_{max} = (\sqrt{4I}+\sqrt{9I})^2 = (2+3)^2 I = 25I
  2. Imin=(9I−4I)2=(3−2)2I=II_{min} = (\sqrt{9I}-\sqrt{4I})^2 = (3-2)^2 I = I
  3. Imax−Imin=24II_{max} - I_{min} = 24I
  4. So x=24x = 24.
Report a mistake

More questions on Superposition, Coherence & Intensity