Diffraction Grating: Possible variables for Experiment

[KDPhysics]

Homework Statement

Devise an experiment investigating the properties of a CD diffraction grating (Can't change wavelength).

Relevant Equations

Diffraction Grating equation for maxima: $d\sin\theta = n\lambda$

For my High School Physics course, I have been tasked to design an experiment investigating the properties of a CD diffraction grating, and we MUST make a graph. Unfortunately, we only have two lasers of different wavelength, so changing the wavelength and measuring $theta$ would be a bad idea.

I have come up with two ideas, one which is theoretically possible, and one which I am not so sure about.

Idea 1: changing the distance from the CD to the viewing screen (which we will call $L$)
We use the diffraction grating equation for maximum interference: $d\sin\theta = n\lambda$ where d is the distance between the bumps on the CD, and define the distance from the laser to the screen L, the angular distance from the principal maxima to the nth order maxima by $theta$, and their distance by $y$.
$$\begin{align}
d\sin(\arctan(\frac{y}{D})) &= n\lambda\\
\arctan(\frac{y}{D}) = \arcsin(n\frac{\lambda}{d})\\
\frac{y}{D} = \tan(\arcsin(n\frac{\lambda}{d}))\\
y = \tan(\arcsin(n\frac{\lambda}{d})) D
\end{align}$$
I can then vary the distance D and measure y for the first order maxima ($y = \tan(\arcsin(\frac{\lambda}{d})) D$). plot them and use the slope of the regression fit to find the distance between the bumps. From there I can calculate the number of bits the CD can hold, and maybe compare with a DVD.

Is the mathematics correct? I am afraid that I may have forgotten some special conditions that make this formula incorrect.

Idea 2: changing the distance between two CDs acting as double diffraction gratings
I place two CDs which act as double diffraction gratings. I change the distance between them, and measure the distance $y$.
Unfortunately, I have no idea if there will be any noticeable difference if i add this second CD, let alone how to develop the mathematics and physics behind this experiment.

 

[haruspex]

What is D? Same as L? What distance is y?
I would have thought that with a laser source the distance from there to the CD was irrelevant.
What about the angle of the beam to the CD?

 

[gneill]

Are you planning to use the CD as a reflective diffraction grating, or "convert" the CD to a transmission grating? Have a look here:

https://www.nnin.org/sites/default/...NG_CDs_AND_DVDs_AS_DIFFRACTION_GRATINGS_0.pdf

 

[gneill]

By the way, note that:
$$\tan(\arcsin(x))) = \frac{x}{\sqrt{1 - x^2}} ~~~~~ \text{where } x < 1 $$

[EDIT] Fixed the equation to reflect the fact that the argument of arcsin function is not an angle. I don't know what I was thinking (or even thinking at all) before..

 

[haruspex]

By the way, note that:
$$\tan(\arcsin(\theta))) = \frac{\theta}{\sqrt{1 - \theta^2}} ~~~~~ \text{where } \theta < 1 \text { is in radians}$$
One radian corresponds to about 57.3° .

Surely the argument to arcsin is not usually an angle. Why would it be in radians?

 

[gneill]

Surely the argument to arcsin is not usually an angle. Why would it be in radians?

Whoops! That's a brain fart on my part. Should have used x instead of θ, and specified that x < 1. I'll edit the post. Sorry about that!