2
$\begingroup$

Recently, I've looked at how the index of refraction of a piece of glass can be related to the angle of rotation from normal incidence and the associated number of fringe transitions through a Michelson interferometer. I'm having a bit of trouble going through the derivation of the following result:

$$ n_g = \frac{(2tn_a - N \lambda_0)(1- \cos{\theta}) + \frac{N^2 \lambda_0^2}{4t}}{2tn_a(1-\cos{\theta}) - N \lambda_0}.$$

In this expression, $n_g$ is the index of refraction of glass, $n_a$ is the index of refraction of air, $t$ is the thickness of the piece of glass, $N$ is the number of fringes, and $\theta$ is the angle of rotation of the glass plate from the perpendicular (see the figure).

The figure for carrying out this analysis is given here:

Figure for the analysis. The solid rectangle is the initial placement of the glass plate, while the dotted rectangle is the rotated plate.

I've almost figured out the whole derivation, but I'm getting stuck at a final step. I'm following the analysis that you can get from Light Principles and Experiments, by George S. Monk (page 377 or thereabout). I'll go through the work I've done so far, and where I'm having trouble. The heart of the problem is that we're looking at the optical path difference between two arms in an interferometer, and that optical path difference is all from this change in distance due to rotating the glass plate.

First, note that the path length when the light travels along $OP$ and the glass plate isn't inclined is given by $n_g \overline{ab} + n_a \overline{bc}$. Then, the optical path as the glass plate is rotated at an angle $\theta$ is given by $n_g \overline{ad} + n_a \overline{de}$. The total increase in optical path length for the interferometer is then twice the difference between these optical path lengths (since the light travels it twice). Therefore, the total increase in optical path length is:

\begin{equation} \label{eq:rotating-interference} \Delta S = 2 \left( n_g \overline{ad} + n_a \overline{de} - \left( n_g \overline{ab} + n_a \overline{bc} \right) \right). \end{equation}

From the figure, some simplifications can be made. Note that $\overline{ad} = t / \cos{\alpha}$. Furthermore, one can use Snell's equation to find that $\angle dce = \theta$, so this gives:

\begin{equation} \overline{de} = \overline{dc} \sin{\theta} = \left( \overline{fc} - \overline{fd} \right) \sin{\theta}. \end{equation}

Then, note that $\overline{fc} = t \tan \theta$ and $\overline{fd} = t \tan \alpha$. Finally, $\overline{bc} + t = t / \cos{\theta}$. For constructive interference, $\Delta S$ must be equal to $N \lambda$. Making the appropriate substitutions, one gets:

\begin{equation} \frac{n_g t}{\cos{\alpha}} + n_a \left( t \tan \theta \sin{\theta} - t \tan \alpha \sin{\theta} \right) - n_g t - \frac{n_a t}{\cos{\theta}} + n_a t = \frac{N \lambda}{2}. \end{equation}

This is where I'm having trouble. In the text I reference, the author simply gives the following explanation (this in paraphrased in my words).

This can be further simplified by using Snell's equation, $n_g \sin{\alpha} = n_a \sin{\theta}$. The result becomes:

\begin{equation} n_g = \frac{(2tn_a - N \lambda_0)(1- \cos{\theta}) + \frac{N^2 \lambda_0^2}{4t}}{2tn_a(1-\cos{\theta}) - N \lambda_0}. \end{equation}

Unfortunately, I can't seem to get that. I've been able to simplify the equation a bit, but not to this final result. Here's what I've done so far.

First, I note that:

\begin{equation} n_g \sin{\alpha} = n_a \sin{\theta} \implies \sin{\alpha} \sin{\theta} = \frac{n_g}{n_a} \sin^2{\alpha}. \end{equation}

This means:

\begin{equation} \frac{n_g}{\cos{\alpha}} - n_a \tan{\alpha} \sin{\theta} = \frac{1}{\cos{\alpha}} \left( n_g - n_a \frac{n_g}{n_a} \sin^2{\alpha} \right) = n_g \cos{\alpha}. \end{equation}

Similarly:

\begin{equation} \tan\theta \sin\theta - \frac{1}{\cos\theta} = -\cos\theta. \end{equation}

Therefore, the equation for constructive interference simplifies to:

\begin{equation} n_g \cos\alpha + n_a \left(1 - \cos\theta \right) = \frac{N \lambda}{2t}. \end{equation}

This is where I'm at right now. I don't see how I can get this to the result shown at the beginning of the post. It seems like I would maybe have to square the equation, but honestly I'm not sure. Any help in figuring out this last step (or series of steps) would be appreciated.

One last thing: using Snell's equation, we can write:

\begin{equation} \cos\alpha = \frac{\sqrt{n_g^2 - n_a^2 \sin^2 \theta}}{n_g}. \end{equation}

Edit: I've worked on the expression a bit more, but still can't quite get the result. Isolating for $n_g \cos\alpha$, we get:

\begin{equation} n_g \cos\alpha = \frac{N \lambda}{2t} - n_a \left(1 - \cos\theta \right). \end{equation}

Squaring this equations and noting that $\left( n_g \cos \alpha \right)^2 = n_g^2 - n_a^2 \sin^2 \theta$, this becomes:

\begin{equation} n_g^2 - n_a^2 \sin^2 \theta = \left( \frac{N \lambda}{2t} \right)^2 - 2 \left( \frac{N \lambda}{2t} \right) n_a \left(1 - \cos\theta \right) + n_a^2 \left(1 - \cos\theta \right)^2. \end{equation}

Then, $\left( 1 - \cos \theta \right)^2 + \sin^2 \theta = 2 \left( 1 - \cos \theta \right)$. Simplifying, we get:

\begin{equation} n_g^2 = \left( \frac{N \lambda}{2t} \right)^2 + 2 n_a \left(1 - \cos\theta \right) \left[ n_a - \frac{N \lambda}{2t} \right]. \end{equation}

I don't know if this helps at all, but I think it might hopefully point us in the right direction.

$\endgroup$
  • 1
    $\begingroup$ Why do you think the index of refraction of the glass changes when you rotate the glass? $\endgroup$ – The Photon Nov 30 '18 at 22:45
  • $\begingroup$ This was a mistake in my characterization. What I meant (and have edited the post to hopefully make clear) is that one can find the index of refraction of glass through an experiment that has you rotating the glass plate by an angle $\theta$ and counting the associated number of fringe transitions $N$. As such, $n_g$ doesn't actually change since $N$ also changes to compensate for it. $\endgroup$ – Germ Nov 30 '18 at 22:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.