AcademyDiffraction
Academy
Single-Slit Diffraction
Level 1 - Physics topic page in Diffraction.
Principle
A single slit produces a broad central maximum and weaker side maxima. Dark fringes occur when wavelets from different parts of the slit cancel in pairs.
Notation
\(a\)
slit width
\(\mathrm{m}\)
\(\lambda\)
wavelength
\(\mathrm{m}\)
\(\theta_m\)
angle of diffraction minimum m
\(\mathrm{rad}\)
\(m\)
minimum order
1
\(L\)
screen distance
\(\mathrm{m}\)
\(y_m\)
screen position of minimum m
\(\mathrm{m}\)
Method
Derivation 1: Pair cancellation
For the first minimum, the path difference between light from the top and bottom of the slit is one wavelength. Rays from the top half then cancel rays from the bottom half.
First minimum
\[a\sin\theta_1=\lambda\]
Minimum order
\[a\sin\theta_m=m\lambda\]
Derivation 2: Screen positions
For small angles, the angular minimum can be converted to a screen position.
Small angle
\[\sin\theta\approx\tan\theta\approx\frac{y}{L}\]
Minimum position
\[y_m\approx\frac{m\lambda L}{a}\]
Derivation 3: Central maximum width
The central maximum extends from the first minimum on one side to the first minimum on the other.
Half-width
\[y_1\approx\frac{\lambda L}{a}\]
Full central width
\[w_{\mathrm{central}}\approx\frac{2\lambda L}{a}\]
Rules
Single-slit minima
\[a\sin\theta_m=m\lambda\]
Small-angle minima
\[y_m\approx\frac{m\lambda L}{a}\]
Central maximum width
\[w_{\mathrm{central}}\approx\frac{2\lambda L}{a}\]
Examples
Question
Light of wavelength
\[520\,\mathrm{nm}\]
passes through a slit of width \[0.250\,\mathrm{mm}\]
Find the first-minimum angle.Answer
\[\sin\theta_1=\frac{520\times10^{-9}}{0.250\times10^{-3}}=2.08\times10^{-3}\]
so \[\theta_1\approx2.08\times10^{-3}\,\mathrm{rad}\]
Checks
- The central maximum is twice as wide as the spacing from the center to the first minimum.
- Narrower slits make wider diffraction patterns.
- The single-slit minimum condition uses \(m=1,2,3,\\ldots\), not \(m=0\).
- The central bright maximum occurs at \(\\theta=0\).