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.)
- A 4
- B 5
- C 3
- D 2
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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