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Physics · Wave Optics

JEE Main 2025 — 23 January, Evening Shift — Question 60

The width of one of the two slits in Young's double slit experiment is d while that of the other slit is xd . If the ratio of the maximum to the minimum intensity in the interference pattern on the screen is 9:49: 4 then what is the value of xx ?

(Assume that the field strength varies according to the slit width.)

  1. Option A:

    2

  2. Option B:

    3

  3. Option C:

    5

    Correct
  4. Option D:

    4

Answer: C

Step-by-step solution

I ∝( width )2\propto(\text { width })^{2}

(I1+I2I1−I2)2=94\left(\frac{\sqrt{I_{1}}+\sqrt{I_{2}}}{\sqrt{I_{1}}-\sqrt{I_{2}}}\right)^{2}=\frac{9}{4}

I1+I2I1−I2=32\frac{\sqrt{I_{1}}+\sqrt{I_{2}}}{\sqrt{I_{1}}-\sqrt{I_{2}}}=\frac{3}{2}

(x+1)d(x−1)d=32\frac{(x+1) \mathrm{d}}{(\mathrm{x}-1) \mathrm{d}}=\frac{3}{2}

⇒3x−3=2x+2\Rightarrow 3 \mathrm{x}-3=2 \mathrm{x}+2

x=5\mathrm{x}=5

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
23 January, Evening Shift
Subject
Physics
Chapter
Wave Optics
Topic
Young's Double Slit Experiment and Its Modifications