# Real world double diffraction grating experiment

I bought a double diffraction grating on Amazon the other day. It says it has 13500 lines per inch. We shone a red laser through it and it made a very nice diffraction grid. The points were spaced at 7.5 degrees.

Experimental details

What I did was measure the distance to a whiteboard with a laser measure (1350mm). I then put the grating right up against the laser measure and mark the diffraction dots on the whiteboard. I was able to measure the distance between the dots. The diagonal of the square grid was 250mm.

This is enough of an angle that the small angle approximations do not apply. So the similar triangles suggest that the wavelength is $\frac{250}{\sqrt{1354^2+250^2}}\left(\frac{25.4}{13500}\right)=342$nm, where 25.4 is the mm/in and the other numbers are as above. The diagram below should make this clear.

If you double this, you get very neatly to 683nm which is pretty close to actual.

Discussion

I took the diagonal because I figure that the diagonal of the diffraction pattern is from the direction parallel to the grid.

The logic here is that the diffraction pattern distance is inversely proportional to the grid spacing, so an increase of $\sqrt{2}$ in the grid spacing (ie the diagonal) causes a decrease of $\sqrt{2}$ in the diffraction pattern. This would fall between the main axis diffraction points and thus make a square. I show this in the diagram below:

On this point, however, I do not entirely understand why the pattern would be such a regular grid. It seems to me that there would be a bunch of next nearest neighbors in off-diagonals creating weak points close in rather than a regular grid.

• I don't know whether this is the case, but in the comments on Amazon, people were also claiming that it is not 13,500 lines per inch. – user1583209 Mar 18 '18 at 13:58
• Please consider writing more useful question titles, see How do we write good question titles? – ACuriousMind Mar 18 '18 at 14:01
• What do you mean by "the points were spaced at 7.5 degrees"? The sine of the angle should be equidistant, not the angle itself, right? – user1583209 Mar 18 '18 at 14:41
• I just measured from the center point to the nearest and next nearest neighbors. – Dr Xorile Mar 19 '18 at 15:57