Strategy for interference problems

[etotheipi]

For instance, consider two sources of light $S_{1}$ and $S_{2}$ where $S_{1}$ is emitting with a phase angle $\frac{\pi}{4}$ greater than $S_{2}$. The light from $S_{1}$ travels a straight distance $d_{1}$ through a medium of refractive index $n_{1}$. The light from $S_{2}$ travels a total distance $d_{2}$ through a medium of refractive index $n_{2}$, reflecting off of a plane mirror as well. Please see the below diagram:

(N.B. I forgot to draw in a barrier, but assume light cannot travel directly from $S_{2}$ to X!)

Let's say we want constructive interference at X where the rays meet. The strategy I came up with is keeping track of the phase angle (modulo an uninteresting $2\pi$) of each light ray at points along the path of each. So I would say that if we set the phase of $S_{1}$ to $\frac{\pi}{4}$ at some constant time, the phase angle at X on the first light ray would be

$\phi_{1} = \frac{\pi}{4} + 2\pi\frac{OPL_{1}}{\lambda}$ (where $\frac{OPL}{\lambda} = \frac{n_{1}d_{1}}{\lambda}$ is the number of wavelengths along the path).

whilst the phase at X on the second wave would be

$\phi_{2} = 0 + 2\pi\frac{OPL_{2}}{\lambda} + \frac{\pi}{2}$ (with the final term due to the reflection).

Finally, I'd set the difference to be a multiple of $2\pi$ and would go from there to solve the system. I can't help but think this is a really convoluted way of going about the question, and I was wondering if anyone had a better method? Thank you in advance!

 

[Charles Link]

The electric fields will add as vectors, and the way you have drawn it, the electric field vectors will be at nearly right angles to each other. The best I can tell, what you are trying to do will not work.

 

[kdv]

For instance, consider two sources of light $S_{1}$ and $S_{2}$ where $S_{1}$ is emitting with a phase angle $\frac{\pi}{4}$ greater than $S_{2}$. The light from $S_{1}$ travels a straight distance $d_{1}$ through a medium of refractive index $n_{1}$. The light from $S_{2}$ travels a total distance $d_{2}$ through a medium of refractive index $n_{2}$, reflecting off of a plane mirror as well. Please see the below diagram:

(N.B. I forgot to draw in a barrier, but assume light cannot travel directly from $S_{2}$ to X!)

View attachment 255057

Let's say we want constructive interference at X where the rays meet. The strategy I came up with is keeping track of the phase angle (modulo an uninteresting $2\pi$) of each light ray at points along the path of each. So I would say that if we set the phase of $S_{1}$ to $\frac{\pi}{4}$ at some constant time, the phase angle at X on the first light ray would be

$\phi_{1} = \frac{\pi}{4} + 2\pi\frac{OPL_{1}}{\lambda}$ (where $\frac{OPL}{\lambda} = \frac{n_{1}d_{1}}{\lambda}$ is the number of wavelengths along the path).

whilst the phase at X on the second wave would be

$\phi_{2} = 0 + 2\pi\frac{OPL_{2}}{\lambda} + \frac{\pi}{2}$ (with the final term due to the reflection).

Finally, I'd set the difference to be a multiple of $2\pi$ and would go from there to solve the system. I can't help but think this is a really convoluted way of going about the question, and I was wondering if anyone had a better method? Thank you in advance!

This is indeed correct and is the only way to do it. If it seems very convoluted, it is just that it is new. The only way to shorten it is to work directly with the difference of the phases at X (which is all that matters), but that really does not change all your reasoning.