Question

In a double-slit experiment, instead of taking slits of equal width, one slit is made twice as wide as the other. Then, in the interference pattern,

• A. the intensities of both, maxima and the minima, will increase.
• B. the intensity of the maxima will increase and minima will have zero intensity.
• C. the intensity of the maxima will decrease and that of minima will increase.
• D. the intensity of the maxima will decrease and the minima will have zero intensity.

Answer 1

The correct option is A the intensities of both, maxima and the minima, will increase.

The resultant intensity at maxima is,

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

And the resultant intensity at minima is,

$I_{min}=(\sqrt{I_1}-\sqrt{I_2}{)}^2$

Also, the intensity of any slit is directly proportional to its width.

When slits of equal width are taken, then $I_1=I_2$ so that intensity at maxima is $4I$ and at minima it is zero.

When width of one slit is doubled, then intensity at maxima will increase as intensity of one of the slits has been increased.

The intensity of minima will also increase (i.e. it will become greater than zero), as now $I_1\ne I_2$, so the resultant intensity at the minima will be greater than zero.

$\begin{array}{c}\text{Why this question?}\\ \text{Note :- When the length or gap of the two slits are unequal, the intensity at the minima will not be zero.}\end{array}$