Diffraction
Diffraction occurs when a wave:
- passes an edge
- passes through a narrow gap
- goes past an object
None of the properties of a wave are changed by diffraction. The wavelength, frequency, period and speed are the same before and after diffraction.
The only change is the direction in which the wave is travelling.
When a wave passes through a gap the diffraction effect is greatest when the width of the gap is about the same size as the wavelength of the wave.
'Double-slit' experiment
For any kind of wave, an interference pattern can be produced in a 'double-slit' experiment. Diffraction results in the two slits acting as two coherent sources.
Slits \(S_{1}\) and \(S_{2}\) are coherent sources. A wave detector may be moved across the interference pattern to find points of constructive and destructive interference.
For visible light, a screen is used to show the interference pattern at the plane of the screen.
Constructive interference
Constructive interference occurs at points where a wave from \(S_{1}\) arrives in phase with a wave from \(S_{2}\).
For this the path difference from \(S_{1}\) and \(S_{2}\) to the point must be a whole number of wavelengths.
Path difference \(=n\lambda\) where \(n\) is an integer.
Destructive interference
Destructive interference occurs at points where a wave from \(S_{1}\) and a wave from \(S_{2}\) arrive out of phase.
For this the path difference must be an odd number of half wavelengths.
Path difference \(=\left(n+\frac{1}{2}\right)\lambda\) where \(n\) is an integer.
Question
Microwave radiation of wavelength \(0.035 m\) is incident on a metal plate which has two slits, \(P\) and \(Q\), as shown.
A microwave detector moved from \(R\) to \(S\) detects a series of maxima and minima.
\(PT=1.30m\) and \(QT =1.16m\)
What type of interference occurs at point \(T\)? You must justify your answer by calculation.
First find the path difference.
\(PT= 1.30m\)
\(QT = 1.16m\)
Path difference \(=PT-QT\)
\(=1.30-1.16\)
\(= 0.14 m\)
Now find out how many wavelengths this path difference is.
Path difference \(= 0.14 m\)
\(\lambda=0.035m\)
\(number\,of\,wavelengths=\frac{path\,difference}{wavelength}\)
\(=\,\frac{0.14}{0.035}\)
\(= 4\)
Now compare with the relationships above.
As path difference is a whole number of wavelengths, constructive interference occurs at point \(T\).