AcademyInterference
Academy
Intensity in Interference Patterns
Level 1 - Physics topic page in Interference.
Principle
Interference intensity depends on phase difference because the electric-field amplitudes add as vectors before intensity is calculated.
Notation
\(I_0\)
intensity from one source alone
\(\mathrm{W\,m^{-2}}\)
\(I\)
resultant intensity
\(\mathrm{W\,m^{-2}}\)
\(\delta\)
phase difference between waves
\(\mathrm{rad}\)
\(E_0\)
field amplitude from one wave
\(\mathrm{V\,m^{-1}}\)
\(V\)
fringe visibility or contrast
1
Method
Derivation 1: Equal amplitudes
Two equal sinusoidal fields with phase difference \(\\delta\) add to a resultant amplitude.
Resultant field amplitude
\[E_{\mathrm{net},0}=2E_0\cos\left(\frac{\delta}{2}\right)\]
Intensity
\[I=4I_0\cos^2\left(\frac{\delta}{2}\right)\]
Derivation 2: Unequal intensities
If the two beams have different intensities, the interference term is smaller than the maximum possible cancellation.
General intensity
\[I=I_1+I_2+2\sqrt{I_1I_2}\cos\delta\]
Maximum
\[I_{\max}=\left(\sqrt{I_1}+\sqrt{I_2}\right)^2\]
Minimum
\[I_{\min}=\left(\sqrt{I_1}-\sqrt{I_2}\right)^2\]
Derivation 3: Visibility
Fringe contrast measures how clear the pattern is.
Visibility
\[V=\frac{I_{\max}-I_{\min}}{I_{\max}+I_{\min}}\]
Rules
Equal-beam intensity
\[I=4I_0\cos^2\left(\frac{\delta}{2}\right)\]
Phase from path
\[\delta=\frac{2\pi\Delta r}{\lambda}\]
Unequal-beam intensity
\[I=I_1+I_2+2\sqrt{I_1I_2}\cos\delta\]
Examples
Question
Two equal beams each have intensity \(I_0\). Find the resultant intensity for
\[\delta=0\]
Answer
\[I=4I_0\cos^2(0)=4I_0\]
Checks
- Equal beams can produce zero intensity at destructive minima.
- Unequal beams cannot cancel completely.
- Maximum intensity from two equal beams is \(4I_0\), not \(2I_0\), because amplitudes add.
- Phase can come from path difference, reflection phase shifts, or source phase.