MCQ
In the Young's double slit experiment using a monochromatic light of wavelength $\lambda$, the path difference (in terms of an integer $n$ ) corresponding to any point having half the peak intensity is :
  • A
    $(2 n+1) \frac{\lambda}{2}$
  • $(2 n+1) \frac{\lambda}{4}$
  • C
    $(2 n+1) \frac{\lambda}{8}$
  • D
    $(2 n+1) \frac{\lambda}{16}$

Answer

Correct option: B.
$(2 n+1) \frac{\lambda}{4}$
b
For half of maximum intensity $2 I _0= I _0+ I _0+2 I _0 \cos \theta$

$\theta$ (Phase difference) $=\frac{\pi}{2}, \frac{3 \pi}{2}, \frac{5 \pi}{2} \ldots \ldots .$.

Path difference is $\frac{\lambda}{4}, \frac{3 \lambda}{4}, \frac{5 \lambda}{4} \ldots \ldots . .\left(\frac{2 n+1}{4} \lambda\right)$

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