Optimizing Phase Shift for Two Waves with Different Wavelengths

[opticaltempest]

I am working on the following problem.

http://img244.imageshack.us/img244/9777/homeworkun4.jpg

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

 

[opticaltempest]

Does anyone have any suggestions?

 

[Doc Al]

Hint: Since \lambda_b > \lambda_a,

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

 

[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)

Adding both equations gives me

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.

 

[Doc Al]

Reread my hint and correct these equations:

Next, I let

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

and

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

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)

 

[opticaltempest]

Since \lambda_b > \lambda_a we should have

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

and

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

 

[opticaltempest]

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

\frac{4L}{\lambda_a}-\frac{4L}{\lambda_b}=1-0.5 \implies4L\bigg(\frac{1}{\lambda_a}-\frac{1}{\lambda_b}\bigg)=0.5 \implies4L\bigg(\frac{\lambda_b - \lambda_a}{\lambda_a \lambda_b}\bigg)=\frac{1}{2} \impliesL=\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.

 

[Doc Al]

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.

 

[opticaltempest]

Thanks for the help Doc Al!