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:

the intensity of both the maxima & the minima increase.

The correct answer is Option (1) → the intensity of both the maxima & the minima increase.

When slit are of equal width,

$I_{max}=\left(\sqrt{I_1}+\sqrt{I_2}\right)^2=\left(\sqrt{I}+\sqrt{I}\right)^2$

$=4I_0$

$I_{max}=\left(\sqrt{I_1}-\sqrt{I_2}\right)^2$

$=\left(\sqrt{I}-\sqrt{I}\right)^2=0$

When slits are of unequal width

$I_{max}'=\left(\sqrt{I_1}+\sqrt{I_2}\right)^2$

$=\left(2I_0+I_0+\sqrt{2I_0I_0}\right)$

$=5.83I_0$

$I_{max}'=\left(\sqrt{I_1}-\sqrt{I_2}\right)^2$

$=2I_0+I_0-\sqrt{2I_0I_0}$

$=0.17I_0$

∴ Intensity of both maxima and minima increases.