# Path Difference Due to Angled Incident Light

If light incident on a diffraction grating makes an angle $\alpha$ with respect to the normal to the grating, show how $$m \lambda = d\sin\theta$$ becomes $$m\lambda = d[\sin(\theta - \alpha) + \sin(\alpha)].$$

I have absolutely no idea what to do here. I've looked into the problem and I've gathered that for incident light $\theta_i$ you get $$m\lambda = d[sin(\theta_i + \theta)],$$ but I'm not sure why that is. No resource I've found has given me a good explanation for that.

If I could prove that $\theta = \alpha + \theta_i$, where $\theta_i$ is the zeroth-order line from the angled incident line, then I could easily substitute my values in, but my problems are that a) I don't understand how to prove the expression in my second paragraph and b) I'm not even sure if $\theta = \alpha + \theta_i$ is true.

I'm mostly looking for the theoretical knowledge to help me work through this example. Any help will be greatly appreciated.

• Apologies, question has been edited. d is multiplied by both $sin(\theta - \alpha)$ and $sin(\alpha)$
– Alex
Commented Nov 4, 2014 at 17:56
• At the heart of it, this question is seeing if you understand how a diffraction grating works. Ponder upon momentum. Commented Nov 4, 2014 at 17:59
• Are you saying that the light rebounds "elastically" such that the incident angle (relative to the zeroth-order diffraction line) will be $\theta = \alpha + \theta_i$ as I stated above? Or am I not making any sense whatsoever?
– Alex
Commented Nov 4, 2014 at 18:04
• A diffraction grating works by transferring quanta of momentum along the direction of the grating. The energy of the photon is conserved, the resolved momentum of the photon perpendicular to the grating remains constant, so the free parameter is the takeoff angle of the photon receiving $\pm n$ units of momentum from the grating (leading to the allowable $\pm n$ diffracted orders). Commented Nov 4, 2014 at 20:10

The diffraction grating acts as an array of light sources. Each slit in the grating emits light, and you calculate the fringe positions by calculating the path lengths from all these sources/slits and identifying the points where the phase differences are multiples of $2\pi$.
• Okay, following that I can understand how $m\lambda = d[sin(\theta) + sin(\alpha)]$, but what I don't understand is why the diffracted angle is still at the same angle relative to the normal (compared to when the incident light is normal). That seems to be what this, as well as a few other sources, are suggesting. If is thr case then I know how to finish answering my question, but I still don't understand WHY it's at the same angle.