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

Answer 1

The correct option is C The intensities of both the maxima and the minima increases

Minimum resultant intensity is given by
$I_{min}=(\sqrt{I_1}-\sqrt{I_2}{)}^2$

$I_1$ and $I_2$ are the intensities of two slits

Slit intensity $\propto (slitwidth{)}^2$
for slits of equal width $I_{\left(min\right)}=0$ as $I_1=I_2$

For slits of unequal width $2a$ and $a$

$I_{min}\propto (2a-a{)}^2$ which is non zero

Clearly, the intensity of minima increases.

Maximum resultant intensity is given by
$I_{max}=(\sqrt{I_1}+\sqrt{I_2}{)}^2=4I$

when one of the slit width is made double intensity of that slit will increase 4 times , thereby increasing the resultant intensity also.

Hence correct option is A.