JEE Mains · Physics · STD 12 - 10. Wave optics
Interference pattern is observed at \('P'\) due to superimposition of two rays coming out from a source \('S'\) as shown in the figure. The value of \('I'\) for which maxima is obtained at \('P'\) is: ( \(R\) is perfect reflecting surface)
- A \(I\, = \,\frac{{2n\lambda }}{{\sqrt 3 - 1}}\)
- B \(I\, = \,\frac{{(2n - 1)\lambda }}{{2(\sqrt 3 - 1)}}\)
- C \(I\, = \,\frac{{(2n - 1\,\lambda )\sqrt 3 }}{{4(2 - \sqrt 3 )}}\)
- D \(I\, = \,\frac{{(2n - 1)\lambda }}{{\sqrt 3 - 1}}\)
Answer & Solution
Correct Answer
(C) \(I\, = \,\frac{{(2n - 1\,\lambda )\sqrt 3 }}{{4(2 - \sqrt 3 )}}\)
Step-by-step Solution
Detailed explanation
From the figure straight path \(\mathrm{SP}=2 l\) Reflected path \(\mathrm{SP}=2 l\) sec \(30^{\circ}\) So path difference is \(2 l\left(\sec 30^{\circ}-1\right)\) Also the ray, when reflected by the mirror, suffers a phase change of \(\pi\) So the total difference in phase is…
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