Derivation of the Diffraction Grating Equation
Second order
wavefront
Zero order
wavefront
First order
wavefronts
The diagram on the left shows a section of a diffraction grating. Each slit diffracts the light waves that pass through
it. The superposition of the wavefronts can lead to constructive interference, where wavefronts are reinforced, or
destructive interference, where wavefronts cancel each other out.
This can be demonstrated by the second diagram, which shows an example of some of the different order
wavefronts. The order of the wavefronts corresponds to the order of the principal maxima.
Draw first order wavefronts onto the first diagram (in the plane shown in the second diagram).
The diagram to the left shows a close up of
two slits, P and Q. The separation of the
slits is given by d, the wavelength of the
light is given by λ, and the angle of
diffraction of the beam is given by θ.
λ
Q
Y
θ
d
P
θ
To nth
order
fringe
For the nth order fringe, the wavefront
emitted from P at time t reinforces a
wavefront emitted n cycles earlier from
the adjacent slit Q. Therefore, now, at
time t, what will be the distance from slit
Q of the original wavefront emitted from
Q? This distance is equal to QY.
Using a trigonometric function, find θ in terms of QY and QP.
Insert for QY and QP such that you derive an equation for sin θ in terms of d, n and λ.
Rearrange to make the subject d sin θ, this is the diffraction grating equation!
What is the largest possible angle for θ?
By using this maximum angle, find an equation for the maximum number of orders, nmax
Relate the number of slits per meter, N, to the spacing of the slits, d.
Therefore for a given order and wavelength, as the distance between the slits decreases, what happens to the angle
of diffraction?
How does this relate to the number of slits per meter?
How else can you alter the angle of diffraction?
N.B. fractions of a degree are expressed as decimals or minutes (‘), such that 1˚ = 60’
Single slit diffraction
envelope
2 slits
3 slits
6 slits
10 slits