Activity: Single slit diffraction
Single slit diffraction occurs when a wave passes through a gap in a barrier. Here you will understand how Huygens' construction leads to the single slit diffraction pattern, observe how the diffraction pattern is affected by changing λ and slit width and derive the equation for the first minimum.
Need to know
When light passes through a narrow slit it spreads out, this is called diffraction. The light does not spread out uniformly but forms bright and dark areas as shown below.
The width of the central maximum depends on the size of the slit, small slit gives wide maximum. With the apparatus set up as shown below the relationship between the angle θ and the slit size,a is:
asinθ = λ
According to Huygens’ construction a wavefront can be considered to be made up of a large number of small wavelet sources. The resulting wave at any point in front of the slit is found by adding up all the wavelets as shown below.
The amplitude at P can be found by adding vectors that represent each wavelet, the phase difference between each wavelet depends on the path difference, but this is complicated by the geometry so we will simplify matters by considering a distant point. Here the waves can be considered parallel.
Considering parallel waves we can see that the waves will add in the straight ahead (each small vector represents a wavelet).
But at a small angle to this position the waves will be slightly out of phase.
This has been modelled using GeoGebra in the following simulation.
Try changing the angle and observe the resultant amplitude.
- What is the phase difference between the first and last wavelet vector at the first minima?
- Why is the amplitude of the first maxima so much smaller than then principal maxima (the middle one).
- Observe the effect of changing λ, a and D. Note that the quantities have no units, and the value of the angle is not displayed. This is because to make the animation fit the screen different scales have been applied to different distances. In the real experiment the angles were much smaller.
- Note that on the simulation the intensity distribution is found from the square of the amplitude. This is because I is proportional to A2.
- This simulation doesn't take into account the change of intensity when the slit width is changed. What else should happen to the graph when the slit width is increased?