MCQ
In a Young’s double slit experiment, ${I_o}$ is the intensity at the central maximum and $\beta $ is the fringe width. The intensity at a point $P$ distant $x$ from the centre will be
  • A
    ${I_o}\cos \frac{{\pi x}}{\beta }$
  • B
    $4{I_o}{\cos ^2}\frac{{\pi x}}{\beta }$
  • ${I_o}{\cos ^2}\frac{{\pi x}}{\beta }$
  • D
    $\frac{{{I_o}}}{4}{\cos ^2}\frac{{\pi x}}{\beta }$

Answer

Correct option: C.
${I_o}{\cos ^2}\frac{{\pi x}}{\beta }$
c
(c) Path difference at point $P = \frac{{xd}}{D}$
Phase difference at point $P = \frac{{2\pi }}{\lambda }\frac{{xd}}{D} = \frac{{2\pi x}}{\beta }$
${I_0} = 4{I_1}$, intensity at point $P$
$I = {I_1} + {I_1} + 2{I_1}\cos \frac{{2\pi x}}{\beta } = 2{I_1}\,\left[ {1 + \cos \frac{{2\pi x}}{\beta }} \right]$
$ = {I_0}{\cos ^2}\frac{{\pi x}}{\beta }$

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