# Homework Help: Intensity in double slit interference

1. Mar 21, 2014

### physiks

I'm a little confused about a small detail when finding the intensity over the screen, as some notes I'm using happen to calculate the same thing twice, with a slight difference the second time.

Each method does the following
Call the electric field at any point E so we have
E=E0ei(kx1-wt)+E0ei(kx2-wt) with x1,x2 the paths from each of the two slits to a point on the screen.
Write each path in terms of a common path x starting midway between the slits, so x1 and x2 both have a path length difference of δ/2 relative to x. Let x1 be the shorter path.
Then x1=x-δ/2 and x2=x+δ/2, and
E=E0ei(kx-kδ/2-wt)+Eoei(kx+kδ/2-wt).
Factoring out the common terms gives
E=E0ei(kx-wt)[2cos(kδ/2)].

This is where this issue is. First it says that the intensity is proportional to E*E whilst later it says it is proportional to E*E/2.

This would give either
4E02cos2(kδ/2)->I=4I0cos2(kδ/2), or
2E02cos2(kδ/2)->I=2I0cos2(kδ/2).
Now the first of these is the expression I always see, so must be right.

I'm a little confused by this. I feel like I might be missing something with regards to time averaging the time varying term as in the E*E/2 method. Can somebody clear this up, thanks :)

Last edited: Mar 21, 2014
2. Mar 21, 2014

### ehild

The intensity is proportional to both E2 and E2/2 .The intensity of the incident light is I0= K E02. The intensity of the diffracted light is I=K [2cos(δ/2)E0]2=[2cos(δ/2)]2 KE02, that is I=4cos2(δ/2) I0.

ehild

3. Mar 21, 2014

### physiks

Why not:

The intensity is proportional to both E2 and E2/2 .The intensity of the incident light is I0= K E02. The intensity of the diffracted light is I=0.5K [2cos(δ/2)E0]2=0.5[2cos(δ/2)]2 KE02, that is I=2cos2(δ/2) I0.

4. Mar 21, 2014

### ehild

In that case, I0=0.5K E02. Be consistent.

ehild

5. Mar 21, 2014

### vela

Staff Emeritus
In the second line, you replaced $E_0^2$ by $I_0$, but it should be $2(E_0^2/2) = 2 I_0$.