Week 13: Interference and Diffraction 429
kr − tω
Etot Eo
kr − tω
Eo
δ
δ/2
δ/2
Figure 179: Phasor diagram for the addition of the electric field components of two slits.
We don’t actually care about thefield strength, of course – we care about theintensity.
The time-averaged intensity of light from asingleslit at the pointPis:
I 0 =^1
2 μ 0 c
|E 0 |^2 (1025)
(from the Poynting vector, as we have seen many times at this point). The total intensity
from the pair of slits is therefore:
Itot= 4I 0 cos^2 (δ/2) (1026)
as you should show, filling in the missing steps.
While this is the completely general solution for the two slit problem (within the ap-
proximations made above) we are often most interested in finding the specific anglesθ
where the interference ismaximumand/orminimum. Clearly theminimaoccur where
cos^2 (δ/2) = 0, which are the phase angles:
δ/2 =±π/ 2 ,± 3 π/ 2 ,± 5 π/ 2 , ... (1027)
or
δ=
2 πd
λ
sin(θ) =±(2m+ 1)π (1028)
or the actual anglesθwhere:
dsin(θ) =±
2 m+ 1
2
λ (1029)
The intensity is zero at the minima.
The maxima occur at the angles where:
δ/2 = 0,±π,± 2 π... (1030)
or
δ/2 =
2 πd
2 λ
sin(θ) =mπ (1031)