Aperture size of a diffraction grating

[truth is life]

Homework Statement

From the intensities of the first-order peaks in the spectrum of a diffraction grating with monochromatic (633 nm) light shining on it, estimate the aperture size. Also known: grating constant (20 nm), distance from grating to screen (25 cm), positions of first-order peaks (plus and minus 8000 microns).

Homework Equations

None, that's what's so frustrating. The book hardly even covers the subject, and certainly doesn't talk about finding the size of the actual lines or whatever in a grating. Online sources are just as bad, usually just mentioning that it is possible, but not giving any details. Trying to do optics problems without someone else there to talk you through them is often an exercise in frustration.

The Attempt at a Solution

None, I have no idea of even where to start as noted above.

 

[Redbelly98]

Are you sure it's 20 nm? The grating spacing must be greater than λ in order to have diffraction peaks other than 0th order.

That being said, my best guess is that they are talking about a transmission grating composed of evenly spaced, narrow slits. If that is the case, then the diffraction pattern consists of bright lines as given by the diffraction grating equation, with intensity multiplied by the single slit intensity pattern. Look up the equation for single slit diffraction intensity, that seems to be what is needed here.

Hope that helps.

 

[truth is life]

Are you sure it's 20 nm? The grating spacing must be greater than λ in order to have diffraction peaks other than 0th order.

Well...fairly sure. (It's a lab, you see, so I'm basing this off of our collected data...which might have mistakes!) I used the equation (can't quite remember how to do latex here)
a*sin theta = m*lambda

lambda is definitely 633 nm (we're using a HeNe laser), m is one. So that means we have
a = lambda/sin theta

Theta is the angle between the central axis of the beam and the radial vector out to whatever point on the screen. So we have

theta = arctan P/D

P being the distance from the center of the beam on the screen to whatever point on the screen (which happens to be mainly what we measured) and D being the distance between the screen and the grating. So a= lambda/sin(arctan(P/D)) = 633 nm/sin(arctan(8000 microns/25 cm)) = 1.979*10^-5 m or ~20 microns. I see what I did, I multiplied by the sin(arctan term when I should have divided. Thanks for catching that!

That being said, my best guess is that they are talking about a transmission grating composed of evenly spaced, narrow slits. If that is the case, then the diffraction pattern consists of bright lines as given by the diffraction grating equation, with intensity multiplied by the single slit intensity pattern. Look up the equation for single slit diffraction intensity, that seems to be what is needed here.

Hope that helps.

Oh yes. That is indeed what they are talking about, and the book (Hecht) very annoyingly never states that the intensity in that case is governed by the single-slit intensity function, which is not at all obvious (I was initially thinking it was modulated by the multi-slit function--logical, right, the whole way a transmission diffraction grating works is a huge number of very narrow slits, after all--which would have been far more trouble).