# Why does fringe width in double slit experiment remain constant if slits get narrower?

The fringe width produced on a screen remains constant as the two double slits get narrower, but you are able to see more fringes on screen.

I don't understand why the fringe width would remain at the same width when the slits get narrower.

In the single slit experiment, fringe width is directly proportional to wavelength/slit width, so I thought that by decreasing slit width (making them narrower), the fringe width would increase. Since the double slit is just the interference of two single slits, why does this not apply there too?

• I'm assuming the distance between the two slits is not changed in your scenario. Only the slit widths are changed. Is that correct? Jun 9, 2020 at 9:03

If I'm misunderstanding anything about your question, feel free to tell me.

As I derived in this post, the (one-dimensional) formula for the double-slit intensity (in the Fraunhofer regime) is

$$I(\theta) = I_{0} \operatorname{sinc}^{2}\left(\tfrac{\pi a\sin\theta}{\lambda}\right) \cos^{2}\left(\tfrac{\pi d\sin\theta}{\lambda}\right)$$

where $$a = \text{slit widths}$$, and $$d = \text{distance between the centers of the two slits}$$.

The double-slit pattern has two patterns at play here: the large envelope pattern (blue graph in the first image) due to single-slit diffraction of each slit, and the small humps within the large pattern due to interference between the two slits (orange graph in the first image).

The slit width $$a$$ is responsible for the width of the large envelope pattern while the slit distance $$d$$ is responsible for the width of the small humps within the pattern.

When you make the slits narrower, you're making $$a$$ smaller, which means you are only changing the envelope pattern (blue graph), not the pattern within the pattern (orange graph).

Hence, the actual fringes don't change size, but how many are displayed within the central spot are changed because the central spot is changing in size.

• what is the meaning of λ and θ in this formula, may I ask? Jun 10, 2020 at 7:18
• @undefined $\lambda$ is the wavelength of the light and $\theta$ is the angle between lines $AB$ and $AC$ where $A$ is where the slits are, $B$ is the middle of the screen, and $C$ is the point on the screen you are interested in. (For example, plugging $\theta=0$ gives the intensity at the middle of the pattern. The formula itself is only valid for the for small angles of $\theta$ which is a reasonable assumption because the screen is supposed to be very far away.) Jun 10, 2020 at 7:24
• @MaximalIdeal But by changing the slith width, you also change the centre to centre distance between the slits. I mean if we narrow the slits more, then we also effectively increase the separation b/w them which means both "a" and "d" change Jun 10, 2020 at 8:33
• thank you, Maximal Ideal Jun 10, 2020 at 11:59

The path difference in the double slit experiment comes from the distance between the two slits that now act as two coherent sources. Of course strictly speaking, the path difference also depends on the width of the slits. But as long as the slit width is much smaller than distance between the two slits these differences are generally not resolved. Pictorially, the maximum discrepancy in the path difference due to finite slit width occurs is the distance between the blue and the green paths. As long as this is much smaller than the distance between the two slits, to a good approximation the interference pattern is slit width independent.