JEE Mains · Physics · STD 12 - 10. Wave optics
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 _______.
- A \(1 / 2\)
- B \(3 / 4\)
- C \(1 / 3\)
- D \(1 / 4\)
Answer & Solution
Correct Answer
(A) \(1 / 2\)
Step-by-step Solution
Detailed explanation
\( \Delta \mathrm{x}=\frac{7 \lambda}{4} \) \( \phi=\frac{2 \pi}{\lambda} \Delta \mathrm{x}=\frac{2 \pi}{\lambda} \times \frac{7 \lambda}{4}=\frac{7 \pi}{2} \) \( \mathrm{I}=\mathrm{I}_{\max } \cos ^2\left(\frac{\phi}{2}\right) \)…
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