11: Wave Phenomena 11.3 Single slit diffraction. Single Slit Diffraction We have already established...
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Transcript of 11: Wave Phenomena 11.3 Single slit diffraction. Single Slit Diffraction We have already established...
Single Slit Diffraction
We have already established that diffraction will occur at an aperture of width b if λ ≥ b
When light is shone at a very thin gap (a ‘slit’), it also diffracts. A diffraction pattern is produced on a screen placed a large distance from the slit:
Explaining The Single Slit Diffraction Pattern
This can be explained if we imagine that all points along the slit width behave as sources emitting secondary wavefronts. (This is called Huygens principle).
If two of these secondary waves meet at a point distant from the slit, they will superimpose upon each other.
b
p = point far from the slit.
p
θ
If these two waves are to superimpose upon each other at p then they must meet. However, we assume that the point is so far away that the two waves are approximately parallel.
First minimum
Darkness is formed. The two waves cancel out.
So the bottom one travels ½ λ further than the top one.
sin θ = ½ λ = λ ½ b b
If θ is very small then sin θ ≈ θ so...
b θ
θ
To first minimum
½ λ
θ = λ b
( Extra note:
We know that for appreciable diffraction to take place, λ ≥ b. If λ = b then sinθ=1 so θ would be 90 degrees.
i.e. There are no subsidiary maxima.
This highlights that diffraction is the spreading out of the wave and does not necessarily show interference maxima and minima. )
Simulations
Effect of λ on θ: Java applet 1
Java applet 2
Java applet 3
Experiment
1. Use the laser and single slit to produce a diffraction pattern.
2. Take measurements (including the ‘half width’ d) and thus determine the slit width b. (λred = 700 nm)
b θd
First minimum
Central maximum
Diffraction from a circular aperture
The angle of the position of the first minimum in this situation is given by the formula...
θ = 1.22 λ b