AcademyDiffraction
Academy
Intensity in Single-Slit Patterns
Level 1 - Physics topic page in Diffraction.
Principle
Single-slit intensity comes from adding phasors from all parts of the slit. The result is a sinc-squared pattern with a strong central maximum and much weaker side maxima.
Notation
\(I\)
intensity at angle theta
\(\mathrm{W\,m^{-2}}\)
\(I_0\)
central maximum intensity
\(\mathrm{W\,m^{-2}}\)
\(\beta\)
single-slit phase parameter
\(\mathrm{rad}\)
\(a\)
slit width
\(\mathrm{m}\)
\(\theta\)
observation angle
\(\mathrm{rad}\)
\(\lambda\)
wavelength
\(\mathrm{m}\)
Method
Derivation 1: Phase spread across the slit
Rays from the top and bottom of the slit have a path difference \(a\sin\theta\), so the phase changes continuously across the aperture.
Full phase spread
\[\Delta\phi=\frac{2\pi a\sin\theta}{\lambda}\]
Half phase parameter
\[\beta=\frac{\pi a\sin\theta}{\lambda}\]
Derivation 2: Add the aperture phasors
The continuous sum of equal wavelets gives an amplitude proportional to \(\\sin\beta/\beta\).
Relative amplitude
\[\frac{E}{E_0}=\frac{\sin\beta}{\beta}\]
Relative intensity
\[\frac{I}{I_0}=\left(\frac{\sin\beta}{\beta}\right)^2\]
Derivation 3: Locate zeros
Intensity is zero when \(\sin\beta=0\) while \(\beta\\ne0\).
Zero condition
\[\beta=m\pi\]
Minima
\[a\sin\theta=m\lambda\]
Rules
Single-slit intensity
\[I=I_0\left(\frac{\sin\beta}{\beta}\right)^2\]
Single-slit parameter
\[\beta=\frac{\pi a\sin\theta}{\lambda}\]
Intensity minima
\[a\sin\theta=m\lambda\]
Examples
Question
Find
\[I/I_0\]
when \[\beta=\pi/2\]
Answer
\[\frac{I}{I_0}=\left(\frac{\sin(\pi/2)}{\pi/2}\right)^2=\left(\frac{2}{\pi}\right)^2=0.405\]
Checks
- The formula is interpreted by its limit at \(\beta=0\), giving \(I=I_0\).
- Minima occur at integer nonzero values of \(\beta/\pi\).
- The side maxima are not equally bright.
- The intensity pattern is symmetric about \(\theta=0\).