JEE Mains · Physics · STD 12 - 10. Wave optics
Width of one of the two slits in a Young's double slit interference experiment is half of the other slit. The ratio of the maximum to the minimum intensity in the interference pattern is _______.
- A \((2 \sqrt{2}+1):(2 \sqrt{2}-1)\)
- B \((3+2 \sqrt{2}):(3-2 \sqrt{2})\)
- C \(9: 1\)
- D \(3: 1\)
Answer & Solution
Correct Answer
(B) \((3+2 \sqrt{2}):(3-2 \sqrt{2})\)
Step-by-step Solution
Detailed explanation
\(\mathrm{I} \propto\) width \(\quad\quad \mathrm{I}_{\max }=\left(\sqrt{\mathrm{I}_1}+\sqrt{\mathrm{I}_2}\right)^2\) \(\therefore \mathrm{I}_1=\mathrm{I}_0, \mathrm{I}_2=2 \mathrm{I}_0 \quad\quad \mathrm{I}_{\min }=\left(\sqrt{\mathrm{I}_1}-\sqrt{\mathrm{I}_2}\right)^2\)…
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