W9_parallel_resonance.eps

Week 13: Interference and Diffraction 427

13.2: Interference from Two Narrow Slits

The first, and simplest, example of interference is monochromatic (constant wavelength)

light falling upon two extremely narrow (slit width less than the wavelength of the light)

separated by a distancedthat is order of a few wavelengths in size. Because the slits are

so close together, they are within the correlation length even of most (monochromatic)

hot sources, so that two slit interference patterns can easily be produced.

To compute the interference pattern produced by two slits, we begin by examining figure

(178), wherein light of fixed wavelengthλfalls normally onto a blocking screen through

which two narrow slits have been cut. Each slit is so narrow that it acts like a “point”

Huygens radiator. Light from one slit (the upper) travels a long distance and falls on a

distant screen. Light from the lower slit travels this distance plus theadditionaldistance

dsin(θ) to arrive at the same point.

d sinθ

d sinθ

d

D

P

θ

θ

θ

θ

λ

r +

r

Figure 178: Two narrow slits act as Huygens radiators when indidentplane wavefronts fall upon
them. Light from the two slits iscoherentandin phaseas it leaves the slits, but arrives atP with
a phase difference that depends on the path difference.

As long as the distanceD between the two slits and the screen is much larger than

dthe distance between the slits themselves then the angleθbetween the horizontal

line shown andbothpaths to the point of observationP is thesame(although this

is not visibly the case in the figure, whereDis not sufficiently large compared tod).

The conditiond≪Dis called theFraunhofer conditionand must be compared to

theFresnel conditionwhich evaluates interference patterns “close to” the slits where

the simplifying Fraunhofer condition does not hold. Fresnel patterns can “easily” be

evaluated as well, but the evaluation requires methodology that is beyond the scope of

this course.

Light from the top slit travels a distancerto arrive at pointP. Light from the bottom slit

travels a distancer+ ∆r=r+dsin(θ) to arrive at the pointP.r≥Danddsin(θ)≤d,

sor≫∆r. We can therefore find the total electric field atPby adding the electric fields

produced by each slit. Let us call the amplitude of the electric field produced by a single