Michelson interferometer -- Plotting experimental data to determine t

[joelkato1605]

Homework Statement

With reference to Eq. 4.6, how would you plot experimental data of total observed wavelength cycles (eg. maxima to maxima) at various incident angles to determine t?

Relevant Equations

See attached images.

See attached images.

 

[Charles Link]

Please show some more introductory material. I don't seem to be able to recognize what the formula refers to.

 

[kuruman]

To joelkato1605:
Yes, plotting $\dfrac{m\lambda_0}{2n_2}$ against $\left[\dfrac{1}{\sqrt{1-\left({\frac{n_1\sin\theta_1}{n_2}}\right)^2}}-1\right]$ will give you a slope (gradient) of $t$.

To @Charles Link:
It looks like there are two refractive media of indices $n_1$ and $n_2$ in one of the arms of the interferometer. The angle of incidence looks like $\theta_1$ and the angle of refraction $\theta_2$. The latter is eliminated from the expression using Snell's law because $\sin\theta_2 =\frac{n_1\sin\theta_1}{n_2}$ which would make the first term in the brackets equal to $\frac{1}{\cos\theta_2}.$

Now if we rearrange the expression a bit, we have $m\lambda_2= \dfrac{2t}{ \cos\theta_2} -2t$ where $\lambda_2$ is the wavelength in medium 2. Well, this looks like to me like we have two identical thin slides of some material of index of refraction $n_2$ in each of the arms of the interferometer. One is oriented at normal incidence and the other at oblique incidence $\theta_1$. The difference in path lengths between the arms for a round trip, as shown on the RHS, is twice the difference between the hypotenuse and the right side of a triangle with right side $t$ and oblique angle of refraction $\theta_2$.

I guess the idea is to find the thickness $t$ by varying the angle of incidence $\theta_1$. The medium of index $n_1$ must be air which cannot be taken equal to 1 because of the sensitivity of the method.

 

[Charles Link]

@kuruman What you have looks almost right, but I'm currently looking at a finer detail at the phase difference for the scenario you presented, and I'm not yet in complete agreement. It would be good to hear from the OP to see if you did indeed get the right scenario.

 

[kuruman]

Yes, of course we should wait to hear from the OP. I too am looking at a finer detail, whether this is a small angle of incidence approximation. I have not examined how applicable this equation is at large angles, e.g. grazing incidence.

 

[Charles Link]

Yes, of course we should wait to hear from the OP. I too am looking at a finer detail, whether this is a small angle of incidence approximation. I have not examined how applicable this equation is at large angles, e.g. grazing incidence.

Perhaps I made an error, but I did the computation of the difference in optical path length of the two arms with the scenario you proposed, and the only way I could get the formula that was given is to compute it incorrectly with what could be an easy error to make. I question whether the OP's text may have made an error.

 

[Charles Link]

To add more detail to the above, using $n_1=1$, I get $\Delta=\frac{2nt}{\sqrt{1-\frac{\sin^2{\theta}}{n^2}}}-\frac{2t}{\sqrt{1-\sin^2{\theta}}}-2(n-1)t$
Edit: I found this to have an error on my part. Upon revising, I get the second term to be $2 t \cos(\theta-\theta_1)/\cos(\theta_1)$, (where $\theta_1$ is the angle in the material), which is also not $2t$.

Perhaps the textbook's way of computing this was to take the difference in the optical path distances inside the two samples=if so, I think that is incorrect. The path length differences outside the sample need to be included. Meanwhile, the change in optical path distance outside the tilted sample is slightly tricky. I think I got it right on my second attempt, as noted in the "Edit" above.

@kuruman You might want to try this calculation, and see if you concur.

 

[kuruman]

@kuruman You might want to try this calculation, and see if you concur.

I am still working on the refinements of my derivation which may very well become what you have. Stay tuned.

 

[kuruman]

@kuruman You might want to try this calculation, and see if you concur.

We differ in the middle term. Why is yours what it is and not simplified to $\frac{2t}{\cos\theta}$? Is something missing?

 

[Charles Link]

We differ in the middle term. Why is yours what it is and not simplified to 2tcos⁡θ? Is something missing?

I thought that's what it would be originally, but if you look at the distance the ray needs to travel in air, you will see it is slightly longer, because the ray does not cross right through the distance $t/\cos{\theta}$, but instead refracts in the material. The distance it covers along the arm of length $L$ during this transit of the material is instead the term I computed in the "Edit" above, with a factor of 2 for the two transits.

 

[kuruman]

My middle term differs from yours. I have$$\Delta=\frac{2 t}{\lambda _0}\left[\frac{n_2}{\cos \left(\theta _2\right)}-\frac{n_1 \cos \left(\theta _1-\theta _2\right)}{\cos \left(\theta _2\right)}-\left(n_2-n_1\right)\right].$$I have kept $n_1$ which you set equal to 1 and my $n_2$ is your $n$. My angles of incidence and refraction are, respectively $\theta_1$ and $\theta_2$ and from Snell's law $\sin\theta_2=\dfrac{n_1\sin\theta_1}{n_2}$ and $\cos\theta_2={\sqrt{1-\left({\dfrac{n_1\sin\theta_1}{n_2}}\right)^2}}.$ The drawing from which this is derived is shown below. It is drawn to scale with tilt angle $\theta_1=50^o$, $n_1=1$ and $n_2=1.44$.

The paths for the two arms are the same up to the point of incidence. The relevant horizontal distance over which different things happen in each arm is the magenta segment. Anything outside that magenta segment is the same for the two arms.

In the tilted medium the red segment has length $\text{Red}=\dfrac{t}{\cos(\theta_2)}.$ The magenta segment is $$\text{Magenta}=\text{(Red)}*\sin[90^o-(\theta_1-\theta_2)]=\frac{t\cos(\theta_1-\theta_2)}{\cos(\theta_2)}.$$Note that the magenta segment is always greater than $t$, therefore in the untilted medium arm we have light going length $t$ in the medium $n_2$ and length $\delta=\text{Magenta}-t$ in medium $n_1$. Then $$\Delta=\frac{2n_2 t}{\lambda_0 \cos(\theta_2)}-\frac{2n_2 t}{\lambda_0}-\frac{2n_1}{\lambda_0} \left[\frac{t\cos(\theta_1-\theta_2)}{\cos(\theta_2)}-t\right].$$ OP's equation does not have the third term in the equation above that arises from considering segment $\delta$ in the figure.

The plot below shows $\Delta$ in units of $2t/\lambda_0$ vs angle of incidence $\theta_1$ for my equation (blue line) and OP's equation (brown line). The parameters of the plot were $n_1=1$ and $n_2=1.44$ to match the drawing.

As aside it looks like we may have lost the OP, but I enjoyed doing this even if it doesn't match OP's physical situation..

 

[Charles Link]

@kuruman We are in agreement. I used different subscripts on my angles, but if you read my post more carefully, I think you would find that we do agree. (You also divided by $\lambda_o$, but that is ok).

Meanwhile good job. :) From your graph, it is clear the OP's formula has a substantial error, if you have the scenario correct, and I would have to believe that you do.

We could use some data from the OP of positions of maxima as a function of angle, and see if the graph of maximum number $m$ vs. $\theta$ has the same shape as your blue curve. It could also be interesting to get a number for $n$ of the material, and an independent measurement, and see if the formula that we came up with gives an accurate and precise value for the thickness $t$.

 

[kuruman]

I divided by $\lambda_0$ because my $\Delta$ is meant to be the number of fringes that pass a fixed point on the pattern as the angle turns. Yes, I agree it would be interesting to have more information from the OP on this.