Interference Tip Sheet
Conditions for Interference
Interference phenomena can occur with any type of wave. The condition for constructive interference for in phase sources is when the path difference is an integer multiple of the wavelength:
Δr = mλ
Δr = path difference (also sometimes denoted Δd)
m = 0, 1, 2 …
λ = wavelength
The condition for destructive interference for in phase sources is when the path difference is a half-integer multiple of the wavelength:
Δr = (m + ½)λ
Do not assume that either of the above equations must be true. It is possible for a particular situation to create neither constructive nor destructive interference.
Both Δr and m give students lots of grief. Geometry should be used to calculate Δr. Determine how far the observer is from one source and subtract how far the observer is from the other source. This should be done symbolically when not all locations are known.
Sometimes you must guess the value of m. Start at m = 0 and work your way up until your solution makes sense.Another option is to calculate Δr/λ which should give clues to mand the type of interference. Don’t ask what m is physically. It is a mathematical artifact, not something you can measure. It is 0, 1, 2 …
Thin Film Interference
For light incident perpendicular to a thin film, there are some adjustments that need to be made:
- The path difference, Δr, will be equal to twice the film thickness, t.
- The wavelength in the thin film will be changed by the index of refraction. This is because the speed changes, so the wavelength changes in proportion while the frequency remains constant.
- The number of inversions upon reflection for reflected rays(when there is reflection off of a higher index of refraction) must be counted. If there are an odd number of inversions, then the equations for constructive and destructive interference get reversed.
The equations are as follows:
2t = mλ/n (constructive for even inversions, destructive for odd)
2t = (m + ½)λ/n (destructive for even inversions, constructive for odd)
Effects of Interference
If the two waves are of equal amplitude, constructive interference at a particular location will result in a wave of double the amplitude of a single source. This gives four times the intensity (brighter light, louder sound, etc.) since I α A2. Two waves of equal amplitude with destructive interference at a particular location will result in zero amplitude and zero intensity (no light, no sound, etc.)
Interference in Everyday Life
Why aren’t these phenomena regularly and easily observed in everyday life?
1)If the wavelength is small, the regions of constructive and destructive interference are small.
2)Most sources of waves do not emit a single wavelength continuously.
3)Reflection off boundaries can make the effective number of sources large.
4)Significant differences in source power or significant differences in path length can result in significantly different amplitudes of the interfering waves. This prevents the destruction from being complete.
Examples of Interference Problems
Example 1
An observer is 9.0 m from speaker A and 3.0 m from speaker B. The speakers are 7.0 m apart. The wavelength is 4.0 m. Specify if there is constructive interference, destructive interference, or neither at the observer’s location.
Solution
Calculate the path difference:
Δr = 9.0 – 3.0 = 6.0 m
Take the ratio Δr/λ to determine type of interference (and possibly m):
Δr/λ = 6.0/4.0 = 1.5
The path difference is a half-integer multiple of the wavelength. This shows that the interference is destructive (m = 1).
Example 2
All things are on the x axis. An observer is at a location of 1.0. Speaker A is at -4.0, and speaker B is at 9.0. The wavelength is 4.0 m. Specify if there is constructive interference, destructive interference, or neither at the observer’s location.
Solution
The observer is 5 meters away from speaker A and 8 meters away from speaker B. This gives the following path difference:
Δr = 8.0 – 5.0 = 3.0 m
Take the ratio Δr/λ to determine the type of interference (and possibly m):
Δr/λ = 3.0/4.0 = 0.75
The path difference is neither an integer multiple of the wavelength nor a half-integer multiple of the wavelength. This shows that the interference is neither constructive nor destructive.
Example 3
All things are on the x axis. An observer is at a location of 6.0. Speaker A is at xa, and speaker B is at 9.0. It is known that 6 > xa > 0. The wavelength is 2.0 m. Find all values of xa to generate constructive interference at the observer’s location.
Solution
The observer is a distance (6.0 – xa) from speaker A and 3.0 meters away from speaker B. This gives the following path difference:
Δr = (6.0 – xa) – 3.0 = 3.0 – xa
Since we don’t know if this will give a positive number for the given range of xa, we need to put an absolute value sign around it:
Δr = |3.0 – xa|
Place this path difference into the condition for constructive interference and solve for all possible values of xa:
Δr = mλ
|3.0 – xa| = mλ
Setting m = 0 gives the following:
|3.0 – xa| = 0
xa = 3.0 meters
Setting m = 1 gives the following:
|3.0 – xa| = 2.0
xa = 5.0 meters or 1.0 meter
Setting m = 2 or more gives answers that are outside the given range specified in the problem statement.
Example 4
An observer is in the middle of a large room with speakers at either end. The speakers emit sound with a wavelength of 3.0 meters.
a) Determine the type of interference at this location
b) Determine how far the observer has to walk towards one speaker to first observe a reversal in the type of interference.
Solution
In the middle of the room (length = L), the observer is L/2 from one speaker and L/2 from the other. Path difference can be calculated as follows:
Δr = L/2 – L/2 = 0
This gives constructive interference (m = 0).
If the observer walks a distance x towards one speaker (and away from the other speaker), the observer is a distance (L/2 + x) from one speaker and (L/2 – x) from the other. This gives a path difference as follows:
Δr = (L/2 + x) – (L/2 – x)
Δr = 2x
Note that this path difference is independent of the size of the room (which is good as this is not given in the problem statement). Use the condition for destructive interference and solve for x:
Δr = (m + ½)λ
2x = (m + ½)λ
x = (m + ½)λ/2
The first location (smallest value of x) is when m = 0:
x = (0 + ½)3/2 = 0.75 meters
Example 5
This is a 2-d problem. An observer is at (0.0, 6.0) m. Speaker A is at (8.0, 0.0) m. Speaker B is at (8.0, 6.0) m. The wavelength is 0.50 m. Specify if there is constructive interference, destructive interference, or neither at the observer’s location.
Solution
The observer is √(6.02 + 8.02) = 10.0 meters from speaker A. The observer is 8.0 meters from speaker B.The path difference is as follows:
Δr = 10.0 – 8.0 = 2.0 m
Take the ratio Δr/λ to determine type of interference (and possibly m):
Δr/λ = 2.0/0.50 = 4
The path difference is an integer multiple of the wavelength (m = 4). This shows that the interference is constructive.
Example 6
All things are on the x axis. An observer is at an unknown location x which is to the right of speaker B. Speaker A is at 2.0 m, and speaker B is at 7.0 m. Calculate some wavelengths for which the observer notes destructive interference.
At first glance, it appears that the path difference cannot be calculated. Have faith and blaze ahead anyway. The person is a distance (x – 2.0) from speaker A and a distance (x – 7.0) from speaker B. The path difference is as follows:
Δr = (x – 2.0) – (x – 7.0)
Δr = 5.0 meters.
Use this path difference in the condition for destructive interference and solve for some possible values of λ:
Δr = (m + ½)λ
λ = Δr/(m + ½)
λ = 5/(0 + ½) = 10 meters
λ = 5/(1 + ½) = 3.3 meters
λ = 5/(2 + ½) = 2.0 meters
etc.
Example 7
White light is incident on a thin film (t = 500 nm) of oil (n = 1.40) floating on water (n = 1.33). Calculate the wavelengths of visible light for which there is constructive and destructive interference.
Solution
Since this is a thin film, the path difference is simply 2t.
Δr = 2t
In order to determine which equations are valid, we must count the number of inversions for the reflected rays. The light that reflects off the oil is inverted because its index of refraction is higher than that of air. The light that reflects off the water is upright (not inverted) because its index of refraction is lower than that of the oil. The total number of inversions is 1, so the conditions for constructive and destructive interference are reversed.
The condition for constructive interference is as follows:
2t = (m + ½)λ/n
Solve the above equation for wavelength, λ.
λ = 2tn/(m + ½)
Now plug in numbers and only accept values for wavelengthsthat are visible:
λ = 2(500 nm)(1.40)/(0 + ½) = 2800 nm (outside visible)
λ = 2(500 nm)(1.40)/(1 + ½) = 933 nm (outside visible)
λ = 2(500 nm)(1.40)/(2 + ½) = 560 nm
λ = 2(500 nm)(1.40)/(3 + ½) = 400 nm
λ = 2(500 nm)(1.40)/(4 + ½) = 311 nm (outside visible)
Wavelengths causing destructive interference can be calculated by removing the ½ from the above equation:
λ = 2tn/m
Now plug in numbers and only accept values for wavelengths that are visible:
λ = 2(500 nm)(1.40)/0 = infinity (nonsense)
λ = 2(500 nm)(1.40)/1 = 1400 nm (outside visible)
λ = 2(500 nm)(1.40)/2 = 700 nm
λ = 2(500 nm)(1.40)/3 = 467 nm
λ = 2(500 nm)(1.40)/3 = 350 nm (outside visible)
Physics is fun!