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JEE Mains · Physics · STD 12 - Wave optics
  • JEE Mains 2024
  • Hard
  • MCQ

In Young's double slit experiment, light from two identical sources are superimposing on a screen. The path difference between the two lights reaching at a point on the screen is \(\frac{7 \lambda}{4}\). The ratio of intensity of fringe at this point with respect to the maximum intensity of the fringe is _______.

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Answer & Solution

Correct Answer

(A) \(1 / 2\)

Step-by-step Solution

Detailed explanation

\( \Delta \mathrm{x}=\frac{7 \lambda}{4} \)

\( \phi=\frac{2 \pi}{\lambda} \Delta \mathrm{x}=\frac{2 \pi}{\lambda} \times \frac{7 \lambda}{4}=\frac{7 \pi}{2} \)

\( \mathrm{I}=\mathrm{I}_{\max } \cos ^2\left(\frac{\phi}{2}\right) \)

\( \frac{\mathrm{I}}{\mathrm{I}_{\max }}=\cos ^2\left(\frac{\phi}{2}\right)=\cos ^2\left(\frac{7 \pi}{2 \times 2}\right)=\cos ^2\left(\frac{7 \pi}{4}\right) \)

\(=\cos ^2\left(2 \pi-\frac{\pi}{4}\right) \)

\(=\cos ^2 \frac{\pi}{4} \)

\(=\frac{1}{2}\)

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