# Energy of twin source interference

1. Apr 8, 2013

### elemis

1. The problem statement, all variables and given/known data

http://www.mediafire.com/view/?3efd7326e49kteb

I've worked out all except part (iv).

These are my workings : http://www.mediafire.com/view/?elswc5g5412zhzq

I just can't see how to get rid of cos(pi*(x1+x2/λ -wt).

Also, where does the I0 come from ?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Apr 8, 2013

### mukundpa

Intensity of a wave, at a point, depends on its amplitude at that point. Which terms of the equation in (ii) represents amplitude at P?

3. Apr 8, 2013

### elemis

I'm guessing A and the second cosine term ? This is just an educated guess, I can really see how or why. Is it to do with the envelope and the carrier wave which are represented by the two cosine terms individually ?

4. Apr 8, 2013

### mukundpa

did that part is time dependent?

5. Apr 8, 2013

What ?

6. Apr 8, 2013

### mukundpa

The first two terms are not changing with time thus they give the amplitude, the third part is time dependent and thus whole equation gives wave displacement at any time t.

7. Apr 8, 2013

### elemis

Okay so taking 2A*first cosine term and squaring it should give me some proportional to Intensity since I = kA^2

Is Io = A^2 then ?

8. Apr 8, 2013

### mukundpa

What is k in the equation and if the distance of P is large what is the path difference in terms of θ ?

9. Apr 8, 2013

### elemis

If P is very very far away then the path difference is dsinθ and k is the wavenumber i.e. 2pi/λ

10. Apr 8, 2013

### mukundpa

Thus the amplitude at P is 2A cos (π d sin θ / λ)

11. Apr 8, 2013

### elemis

I've worked that out already in the workings I attached in the OP.

How do I take this and transform into the equation in part (iv) ?

12. Apr 8, 2013

### mukundpa

I at P is square of this and A2 = I0