JEE Mains · Physics · STD 12 - 10. Wave optics
In a Young's double slit experiment, the distance between the two identical slits is \(6.1\) times larger than the slit width. Then the number of intensity maxima observed with in the central maximum of the single slit diffraction pattern is
- A \(3\)
- B \(6\)
- C \(12\)
- D \(24\)
Answer & Solution
Correct Answer
(C) \(12\)
Step-by-step Solution
Detailed explanation
In diffraction, angular width of central maxima is \(\frac{2 \lambda}{b}\) where \(b\) is slit width. In Young's double slit experiment, angular width of each fringe is \(\frac{\lambda}{d}\) where \(d\) is the distance between the two slits. Given: \(d=6.1 b\) So, number of…
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