# Phase shift between two waves

1. Jan 26, 2008

### opticaltempest

I am working on the following problem.

3. The attempt at a solution

I realize that the relative phase shift between ray 1 and ray 2 will be

$$\frac{4L}{\lambda}$$

Next, I let

$$\frac{4L}{\lambda_a}=1$$

and

$$\frac{4L}{\lambda_b}=1.5$$

I did this in hopes to make the phase shift in wavelengths for $$\lambda_a$$ an integer number and the phase shift for $$\lambda_b$$ an integer plus 0.5 (to put the waves exactly out of phase).

I tried various paths from this point but cannot get a valid length for $$L$$ that puts $$\lambda_a$$ in phase and $$\lambda_b$$ out of phase. Could anyone offer a suggestion on how to proceed?

Thanks

Last edited: Jan 26, 2008
2. Jan 27, 2008

### opticaltempest

Does anyone have any suggestions?

3. Jan 27, 2008

### Staff: Mentor

Hint: Since $\lambda_b > \lambda_a$,

$$\frac{4L}{\lambda_b} < \frac{4L}{\lambda_a}$$.

4. Jan 27, 2008

### opticaltempest

I have two equations with three unknowns so I should be able to solve for one variable in terms of the other two. I use these two equations

$$4L - \lambda_a=0$$ (1)

$$4L - 1.5\lambda_b=0$$ (2)

$$8L-\lambda_a-1.5\lambda_b=0$$

Solving for L gives me

$$L=\frac{\lambda_a+1.5\lambda_b}{8}$$

When testing this equation, I find that I don't get an integer number of wavelengths for the phase shift for $$\lambda_a$$ and an integer + 0.5 wavelengths phase shift for $$\lambda_b$$.

I have also tried a few different paths but still get nowhere.

5. Jan 27, 2008

### Staff: Mentor

Reread my hint and correct these equations:

6. Jan 27, 2008

### opticaltempest

Since $$\lambda_b > \lambda_a$$ we should have

$$\frac{4L}{\lambda_a}=1$$ (1)

and

$$\frac{4L}{\lambda_b}=0.5$$ (2)

7. Jan 27, 2008

### opticaltempest

When I subtract (2) from (1) I get the equation

$$\frac{4L}{\lambda_a}-\frac{4L}{\lambda_b}=1-0.5 \implies$$

$$4L\bigg(\frac{1}{\lambda_a}-\frac{1}{\lambda_b}\bigg)=0.5 \implies$$

$$4L\bigg(\frac{\lambda_b - \lambda_a}{\lambda_a \lambda_b}\bigg)=\frac{1}{2} \implies$$

$$L=\frac{1}{8}\frac{\lambda_a \lambda_b}{\lambda_b-\lambda_a}$$

Say $$\lambda_a=100nm$$ and $$\lambda_b=120nm$$. Using the above equations, we find L to be 75nm. The relative phase shift between the two rays for a 100nm wavelength is

$$\frac{4*75nm}{100nm}=3.0$$

3 wavelengths phase shift puts the 100nm wavelength exactly in phase!

For the 120nm wavelength, the relative phase shift between the rays is

$$\frac{4*75nm}{120nm}=2.5$$

The relative phase shift between the two rays is 0.5 wavelengths and they are exactly out of phase. This appears to be a correct answer. I am hoping this is the least L that satisfies the problem.

Last edited: Jan 27, 2008
8. Jan 27, 2008

### Staff: Mentor

Looks good to me!

The most general way to express the relationship would be:

$$\frac{4L}{\lambda_a}-\frac{4L}{\lambda_b}= n + 0.5$$

The smallest value of L would be when n = 0.

9. Jan 27, 2008

### opticaltempest

Thanks for the help Doc Al!