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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


Doing this at home?

This is difficult to do without a laser and prepared slit, so you'll have to use the same video as in the SL Activity: Light waves.

Experiment

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

Method

Using the Vernier calliper as a slit form a diffraction pattern on the wall. By measuring the y and D find θ and determine the wavelength of the laser.

The PASCO laser has a wavelength 650 nm

  • Try varying the slit width and observe what happens to the pattern.

Warning

The laser is a powerful source of parallel light if you look directly at the beam it will be focused as a very small spot on your retina. This can burn a hole leaving a permanent blind spot.

Using vectors to represent waves

When two waves are incident at a point the total displacement is the vector sum of the individual displacements. When they are in phase they add together, when out of phase they cancel and if the phase difference is some angle φ then the sum is the resultant of two vectors with angle φ between them. This is best shown in an animation.

In this GeoGebra animation two waves are added as their phase is changed.

The two black vectors represent the two waves, the red vector the resultant. Note how the resultant vector is the same as the amplitude of the resultant wave.

Actually that's quite difficult to understand so here's another attempt (it's important that you get this).

The displacement of the green wave is represented by the green rotating vector, the actual displacement at the origin is given by the vertical component of this vector as shown.

The amplitude of the wave is the same as the length of the vector

The purple wave is represented by the purple vector. The phase of this wave can be varied using the slider.

The red wave is the superposition of the green and purple. Notice how the displacement is the vertical component of the resultant of the green and blue vectors.

If we want to find the amplitude we simply add the vectors.

Theory

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?

Deriving the equation

You have seen that the phase difference between the first and last wavelets at the first minima is 2π this means the path difference is λ. We can represent this on a simple diagram.

  • Write an equation for the sin of angle θ

    Summary

    If you would like further explanation of single slit diffraction you can watch this video.

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