Question

In a double slit experiment instead of taking slits of equal widths, one slit is made twice as wide as the other, then in the interference pattern

• A. the intensities of both the maxima and the minima increases
• B. the intensity of the maxima increases and the minima has zero intensity
• C. the intensity of maxima decreases and that of minima increases
• D. the intensity of maxima decreases and the minima has zero intensity

Answer 1

Answer: (A)

In interference we know that
$ I_{max}=(\sqrt{I_1}+\sqrt{I_2})^2$
and $ I_{min}=(\sqrt{I_1}-\sqrt{I_2})^2$
Under normal conditions (when the widths of both the slits
are equal)
$ I_1=I_2=I$ (say)
$\therefore I_{max}=4I \, \, and \, \, I_{min}=0$
When the width of one of the slits is increased. Intensity due
to that slit would increase, while that of the other will remain
same. So, let:
$ I_1=I$
and $ I_2=\eta I (\eta > 1)$
Then, $ I_{max}=I(1+\sqrt {\eta})^2 > 4I$
and $ I_{min}=I(\sqrt {\eta}-1)^2 > 0$
$\therefore $ Intensity of both maxima and minima is increased.