Animated Solution for Physics - Optics: 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

Imagine a standard Young's Double Slit Experiment (YDSE). We have two slits, $S_1$ and $S_2$, and a screen where the beautiful, alternating bright and dark interference pattern forms. The intensity of this pattern is a direct consequence of the superposition of light waves emerging from the two slits.

To understand how the pattern changes, we must look at the fundamental formulas governing the maximum and minimum intensities in an interference pattern. The intensity at any point on the screen depends on the individual intensities of the light from each slit, let's call them $I_1$ and $I_2$.

The maximum intensity, which corresponds to the bright fringes, is given by:

The minimum intensity, which corresponds to the dark fringes, is given by:

Equal Slit Widths

In the classic setup, both slits have identical widths. This means they allow the same amount of light to pass through, resulting in equal intensities: $I_1=I_2=I$.

Substituting this into our formulas, we find:

This is the ideal scenario. The maximum intensity is four times the intensity of a single slit, and the minimum intensity is perfectly zero. This perfect zero means the dark fringes are completely black, providing maximum contrast and incredibly sharp fringes.

Unequal Slit Widths

Now, let's introduce a twist. What happens if we make one slit wider than the other? For instance, the problem states one slit is made twice as wide. The intensity of light passing through a slit is directly proportional to its width. Therefore, the wider slit will emit light of a higher intensity.

Let's denote the new intensities as $I_1=I$ and $I_2=\eta I$, where $\eta >1$ (since one slit is wider).

Let's recalculate the maximum and minimum intensities with these new values: Since $\eta >1$, the term $(1+\sqrt{\eta }{)}^2$ is strictly greater than $4$. Therefore, $I_{\max }^{\prime }>4I$.

Now for the minimum intensity: Again, since $\eta >1$, the term $(\sqrt{\eta }-1{)}^2$ is strictly greater than $0$. Therefore, $I_{\min }^{\prime }>0$.

By comparing the new intensities with the ideal case, we can draw a clear conclusion. The maximum intensity has increased from $4I$ to a value greater than $4I$. Simultaneously, the minimum intensity has increased from $0$ to a value greater than $0$.

Thus, the intensities of both the maxima and the minima increase.

While the bright fringes get brighter, the dark fringes are no longer perfectly dark. They now have some residual brightness. This phenomenon reduces the overall contrast (or visibility) of the interference pattern, making the fringes appear somewhat washed out compared to the ideal case.