You should be familiar with the idea that, when light passes through a slit, it is diffracted (caused to spread out in arcs from the slit). The amount of diffraction increases the closer the slit width is to the wavelength of the light. Consider the animation on the right. Light from a light source is caused to pass through two slits. It is diffracted at both these slits, and so it spreads out in two sets of arcs.

Now, apply superposition of waves to this situation. At certain points, the peaks (or troughs) of the waves will coincide, creating constructive interference. If this occurs on a screen, then a bright 'fringe' will be visible. On the other hand, if destructive interference occurs (a peak coincides with a trough), then no light will be visible at that point on the screen.

## Contents

## Calculating the angles at which fringes occurEdit

If we wish to calculate the position of a bright fringe, we know that, at this point, the waves must be in phase. Alternatively, at a dark fringe, the waves must be in antiphase. If we let the wavelength equal λ, the angle of the beams from the normal equal θ, and the distance between the slits equal d, we can form two triangles, one for bright fringes, and another for dark fringes (the crosses labelled 1 and 2 are the slits):

The length of the side labelled λ is known as the path difference. For bright fringes, from the geometry above, we know that:

Therefore:

However, bright fringes do not only occur when the side labelled λ is equal to 1 wavelength: it can equal multiple wavelengths, so long as it is a whole wavelength. Therefore

,

where n is any integer.

Now consider the right-hand triangle, which applies to dark fringes. We know that, in this case:

We can generalise this, too, for any dark fringe. However, if 0.5λ is multiplied by an even integer, then we will get a whole wavelength, which would result in a bright, not a dark, fringe. So, n must be an odd integer in the following formula:

## Calculating the distances angles correspond to on the screenEdit

At this point, we have to engage in some slightly dodgy maths. In the following diagram, p is path difference, L is the distance from the slits to the screen and x is the perpendicular distance from a fringe to the normal:

Here, it is necessary to approximate the distance from the slits to the fringe as the perpendicular distance from the slits to the screen. This is acceptable, provided that θ is small, which it will be, since bright fringes get dimmer as they get further away from the point on the screen opposite the slits. Hence:

If we substitute this into the equation for the path difference p:

So, at bright fringes:

, where n is an integer.

And at dark fringes:

, where n is an odd integer.

You may have noticed that is not precisely the distance from a fringe to the normal, differing from it by the amount . However, in a diffraction grating, the distance between the slits is negligible in comparison to , so it may be ignored.

## Diffraction GratingEdit

A diffraction grating consists of a lot of slits with equal values of d. As with 2 slits, when , peaks or troughs from all the slits coincide and you get a bright fringe. Things get a bit more complicated, as all the slits have different positions at which they add up, but you only need to know that diffraction gratings form light and dark fringes, and that the equations are the same as for 2 slits for these fringes.

## QuestionsEdit

1. A 2-slit experiment is set up in which the slits are 0.03 m apart. A bright fringe is observed at an angle 10° from the normal. What sort of electromagnetic radiation was being used?

2. Light, with a wavelength of 500 nm, is shone through 2 slits, which are 0.05 m apart. What are the angles to the normal of the first three dark fringes?

3. Some X-rays, with wavelength 1 nm, are shone through a diffraction grating in which the slits are 50 μm apart. A screen is placed 1.5m from the grating. How far are the first three light fringes from the point at which the normal intercepts the screen?