W9_parallel_resonance.eps

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)