JEE Mains · Physics · STD 12 - Wave optics
- JEE Mains 2014
- Hard
- MCQ
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)
Pick an option to check your answer.
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 \(2 l\left(\sec 30^{\circ}-1\right) \times \frac{2 \pi}{\lambda}+\pi\)
For constructive interference
\(2 l\left(\sec 30^{\circ}-1\right) \times \frac{2 \pi}{\lambda}+\pi=2 n \pi\)
Solving this, we get \(l=\frac{(2 n-1) \lambda \sqrt{3}}{4(2-\sqrt{3})}\)
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