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JEE Mains · Physics · STD 12 - 10. Wave optics

The width of one of the two slits in Young's double slit experiment is d while that of the other slit is \(x \mathrm{~d}\). If the ratio of the maximum to the minimum intensity in the interference pattern on the screen is \(9: 4\) then what is the value of \(x\) ?
(Assume that the field strength varies according to the slit width.)

  1. A 4
  2. B 5
  3. C 3
  4. D 2
Verified Solution

Answer & Solution

Correct Answer

(B) 5

Step-by-step Solution

Detailed explanation

\begin{aligned} & \mathrm{I} \propto(\text { width })^2 \\ & \left(\frac{\sqrt{\mathrm{I}_1}+\sqrt{\mathrm{I}_2}}{\sqrt{\mathrm{I}_1}-\sqrt{\mathrm{I}_2}}\right)^2=\frac{9}{4} \\ &…

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