Gratings
A grating consists of many slits, or lines, close together. In a grating, the spacing between adjacent lines is constant.
Gratings are also used to produce interference patterns. When waves are incident on a grating all of the lines act as coherent sources of the waves.
Compared to a double slit, the interference pattern produced by a grating has fewer, more widely spaced points of maximum intensity.
Monochromatic source of light
The diagram above illustrates the effect of a grating on a monochromatic source of light.
The central maximum is also called the zero order maximum.
For other orders of maxima, the relationship between the wavelength of light \(\lambda\) and angle \(\theta\) is shown by the equation:
\(dsin\theta =n\lambda\)
In this equation:
- \(d\) is the spacing of adjacent lines on the grating
- \(n\) is the order number for the maximum
For the case shown in the diagram \(dsin\theta =2\lambda\)
Question
A light source of unknown frequency is incident on a grating.
The grating has \(2.50 \times 10^{5}\) lines per metre.
The third order maximum of intensity is at an angle of \(22^\circ\) to the central maximum.
Calculate the frequency of the light.
First use the grating relationship to find the wavelength.
Grating has \(2.50 \times 10^{5}\) lines per metre.
Line spacing \(d=\,\frac{1}{2.50\times10^5}\)
\(=4\times10^{-6}m\)
\(\theta=22^\circ\)
\(n=3\)
\(\lambda=?\)
\(dsin\theta=n\lambda\)
\(4\times10^{-6}\times sin22^\circ=3\lambda\)
\(\lambda=4.99475\times10^{-7}m\)
Now use wave equation to find \(f\).
\(\lambda = 4.99475 \times 10^{-7}m\)
\(v=3.00\times10^{8}ms^{-1}\)
\(f=?\)
\(v= f\lambda\)
\(3.00\times 10^{8}=f\times 4.99475\times 10^{-7}\)
\(f=6.0063\times 10^{14}Hz\)
Frequency of light \(=6.0\times 10^{14}Hz\)