Double Slit Interference
Two Source Interference
- For two-source interference fringes to be observed, the sources of the wave must be:
- Coherent (constant phase difference)
- Monochromatic (single wavelength)
- When two waves interfere, the resultant wave depends on the phase difference between the two waves
- This is proportional to the path difference between the waves which can be written in terms of the wavelength λ of the wave
- As seen from the diagram, the wave from slit S2 has to travel slightly further than that from S1 to reach the same point on the screen
- The difference in this distance is the path difference
Path difference of constructive and destructive interference is determined by wavelength
- For constructive interference (or maxima), the difference in wavelengths will be an integer number of whole wavelengths
- For destructive interference (or minima) it will be an integer number of whole wavelengths plus a half wavelength
- n is the order of the maxima/minima since there is usually more than one of these produced by the interference pattern
- An example of the orders of maxima is shown below:
Interference pattern of light waves shown with orders of maxima
- n = 0 is taken from the middle, n = 1 is one either side and so on
Young's Double Slit Experiment
- Young’s double-slit experiment demonstrates how light waves can produce an interference pattern
- The setup of the experiment is shown below:
Young’s double-slit experiment arrangement
- When a monochromatic light source is placed behind a single slit, the light is diffracted producing two light sources at the double slits A and B
- Since both light sources originate from the same primary source, they are coherent and will therefore create an observable interference pattern
- Both diffracted light from the double slits create an interference pattern made up of bright and dark fringes
Worked example
Two coherent sources of sound waves S1 and S2 are situated 65 cm apart in air as shown below.The two sources vibrate in phase but have different amplitudes of vibration. A microphone M is situated 150 cm from S1 along the line normal to S1 and S2. The microphone detects maxima and minima of the intensity of the sound. The wavelength of the sound from S1 to S2 is decreased by increasing the frequency.Determine which orders of maxima are detected at M as the wavelength is increased from 3.5 cm to 12.5 cm.
Exam Tip
The path difference is more specifically how much longer, or shorter, one path is than the other. In other words, the difference in the distances. Make sure not to confuse this with the distance between the two paths.