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 :
  • $1 / 2$
  • B
     $3 / 4$
  • C
     $1 / 3$
  • D
     $1 / 4$

Answer

Correct option: A.
$1 / 2$
a
$ \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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