Question

Compute the intensity of light I on a screen at a distance L >> a of two very long (infinite) slits of width a, a distance d (>a) apart.

Answer 1

At one point on the screen we have the following electric filed intensities.

Let the electric filed of the slit 1 be

E1 = E0*sin(w*t)

and the electric field of slit 2 be

E2 =E0*sin(w*t +fi)

where fi is the phase difference between the phase shift.

Total electric field is the sum :

E = E1+E2 =E0*[sin(wt) +sin(wt+fi)] = 2*E0*cos(fi/2)*sin(wt +fi/2)

because sin(a) +sin(b) = 2*sin((a+b)/2)*sin((a-b)/2)

The intensity of ligt is the time average of the square of electric field.

I =<E^2> = 4E0^2*<cos^2(fi/2)>*<sin^2(wt+fi/2)> = 2*E0^2*cos^2(fi/2)

since <sin^2(wt+fi/2)> =1/2

Now phase phi as function of path difference (delta) can be written as

delta/lambda = fi/(2*pi)

and

delta =d*sin(theta)

where theta is the angle between the horizontal and the ray of light emerging from one slit.

tan(theta) = D/L

(D is the distance from central brightest frindge to the interference point)

Therefore

I =I0*cos^2(pi*d*sin(theta)/lambda)

Observation: for L>>a (slit opening width) it only counts the distance between slits d, not the slit opening width. See the figure below.