Refractive index of a prism using a spectrometer

[VVS2000]

In the image attached, How is it that the angle shown is 2A? And what exactly is the position of min deviation? Why does it have to be in the given position as shown?
Thanks in advance!

 

[Cutter Ketch]

In the image attached, How is it that the angle shown is 2A?

Keep in mind that the angle of incidence equals angle of reflection, and you should be able to do the trig yourself,

And what exactly is the position of min deviation? Why does it have to be in the given position as shown?

It doesn’t. They didn’t want to give it away. You are supposed to rotate the table until the beam deviates the least. You will find it does not occur in the orientation shown.

 

[VVS2000]

Keep in mind that the angle of incidence equals angle of reflection, and you should be able to do the trig yourself,
It doesn’t. They didn’t want to give it away. You are supposed to rotate the table until the beam deviates the least. You will find it does not occur in the orientation shown.

Well I was instructed to rotate the table until the image moves in the opposite direction of rotation. Why should this be the point where light deviates the least?

 

[Charles Link]

I went through the derivation of the minimum angle of deviation of the prism spectrometer many years ago in college. If I remember correctly, it is necessary to use an incident angle $\alpha$, and an emerging angle $\beta$, and take partial derivatives, and set them equal to zero. One thing that results is that $\alpha=\beta$ at the minimum angle of deviation. If I did a little digging, I think I could find the lab manual where the formula was derived.

 

[VVS2000]

I went through the derivation of the minimum angle of deviation of the prism spectrometer many years ago in college. If I remember correctly, it is necessary to use an incident angle $\alpha$, and an emerging angle $\beta$, and take partial derivatives, and set them equal to zero. One thing that results is that $\alpha=\beta$ at the minimum angle of deviation. If I did a little digging, I think I could find the lab manual where the formula was derived.

No I was'nt talking about the derivation. When you're doing it practically, why does the angle which you measure in the picture I attached is 2A? Twice the angle of prism?

 

[Charles Link]

No I was'nt talking about the derivation. When you're doing it practically, why does the angle which you measure in the picture I attached is 2A? Twice the angle of prism?

I was trying to answer your second question. I believe the first one is simply trigonometry.

 

[VVS2000]

I was trying to answer your second question. I believe the first one is simply trigonometry.

Oh sry. Yeah I think I got the first one. I thought you were offering a different perspective
Ok, so practically is'nt there anyway to convince that position is min deviation position?

 

[Charles Link]

For the first one, if you use the symmetry, which may not be necessary, the incident angle on one side is $90-A/2$, making the total angle $180-A$ on each side, for a total of $360-2A$. That leaves $2A$ for the remaining part on the circle.

 

[Charles Link]

Deviation angle $\Psi=\Psi(\alpha, n, A)$, if I remember correctly. If you take $\frac{\partial{\Psi}}{\partial{\alpha}}=0$, to find $\Psi_{min}$, I think a little computing gave the result that $\alpha=\beta$ at the minimum. I could look for the lab manual tomorrow, but I don't think the derivation is a very difficult one. We should be able to derive it ourselves with a little effort, using Snell's law, etc. $\\$ Edit: With a little effort, I succeeded in doing the derivation of the formula. I would need to write it up neatly, (drawing a diagram, etc.), and take a photo of the derivation, etc., if you are interested. It is somewhat difficult to derive, but not tremendously so.

 

[VVS2000]

Deviation angle $\Psi=\Psi(\alpha, n, A)$, if I remember correctly. If you take $\frac{\partial{\Psi}}{\partial{\alpha}}=0$, to find $\Psi_{min}$, I think a little computing gave the result that $\alpha=\beta$ at the minimum. I could look for the lab manual tomorrow, but I don't think the derivation is a very difficult one. We should be able to derive it ourselves with a little effort, using Snell's law, etc.

Ok I will give this a try.
Thanks!

 

[Charles Link]

Ok I will give this a try.
Thanks!

See my edit to post 9.

 

[VVS2000]

Deviation angle $\Psi=\Psi(\alpha, n, A)$, if I remember correctly. If you take $\frac{\partial{\Psi}}{\partial{\alpha}}=0$, to find $\Psi_{min}$, I think a little computing gave the result that $\alpha=\beta$ at the minimum. I could look for the lab manual tomorrow, but I don't think the derivation is a very difficult one. We should be able to derive it ourselves with a little effort, using Snell's law, etc. $\\$ Edit: With a little effort, I succeeded in doing the derivation of the formula. I would need to write it up neatly, (drawing a diagram, etc.), and take a photo of the derivation, etc., if you are interested. It is somewhat difficult to derive, but not tremendously so.

Sure! If that's ok with you...

 

[sophiecentaur]

No I was'nt talking about the derivation. When you're doing it practically, why does the angle which you measure in the picture I attached is 2A? Twice the angle of prism?

This sort of question is much better answered by looking at an elementary optics textbook (or searching Google Images - which gives you many pictorial hits). Alternatively, just draw a diagram of a light ray, reflected in a mirror (a simpler model to start with) and see where it goes when you rotate the mirror a bit. If you don't like Maths too much, this should help. OR play with an actual mirror and light from a torch / LED pointer.

 

[sophiecentaur]

No I was'nt talking about the derivation. When you're doing it practically, why does the angle which you measure in the picture I attached is 2A? Twice the angle of prism?

You can't say "why" something like that happens without a 'derivation'. The word may look like intimidating maths and something to steer clear of but the only alternative is to do many practical measurements with different angles and then you may come to the same conclusion that a few lines of maths will give you. (Only you can never be sure that the practical results apply generally and that's why we do sums in Physics)

 

[Charles Link]

 

[Charles Link]

The following will apply to the figure of the above post.
Deviation angle $\Psi=\alpha+\beta-\gamma$.
also $(90-\alpha')+(90-\beta')+\gamma=180$, so that $\gamma=\alpha'+\beta'$.
With Snell's law: $\alpha'=\sin^{-1}(\sin(\alpha) /n)$, so that $\beta'=\gamma-\sin^{-1}(\sin(\alpha )/n) =\sin^{-1}(\sin(\beta)/n)$.
There are just a few more steps.
We have $\beta=\beta(\alpha)$.
$\frac{d \Psi}{d \alpha}=0$ gives $1+\frac{d \beta}{d \alpha}=0$ at the minimum.
For $\beta'$, we have something of the form $F(\beta)=\gamma-F(\alpha)$.
This gives $F'(\alpha) \, d \alpha=-F'(\beta) \, d \beta$.
Since $d\beta=-d \alpha$ at the minimum, we get $F'(\alpha)=F'(\beta)$,
and it should be fair to conclude that $\alpha=\beta$ at the minimum.

 

[Charles Link]

I'm going to complete the derivation using one more post:
Since $\alpha=\beta$ at the minimum, we have
$\Psi_{min}=2 \alpha-\gamma$,
and also $2 \sin^{-1}((\sin{\alpha})/n)=\gamma$,
so that $(\sin{\alpha})/n=\sin(\gamma /2)$,
so that $n=\frac{\sin{\alpha}}{\sin(\gamma /2)}$.
Finally $n=\frac{\sin((\Psi_{min}+\gamma)/2)}{\sin(\gamma /2)}$.

 

[VVS2000]

I'm going to complete the derivation using one more post:
Since $\alpha=\beta$ at the minimum, we have
$\Psi_{min}=2 \alpha-\gamma$,
and also $2 \sin^{-1}((\sin{\alpha})/n)=\gamma$,
so that $(\sin{\alpha})/n=\sin(\gamma /2)$,
so that $n=\frac{\sin{\alpha}}{\sin(\gamma /2)}$.
Finally $n=\frac{\sin((\Psi_{min}+\gamma)/2)}{\sin(\gamma /2)}$.

Oh... ! Now it makes full sense! Thanks man! I Hope to return the favour sometime later

 

[Charles Link]

It may interest you that in the use of a spectrometer, $n=n(\lambda)$, so that the $\beta$ will be different between $\lambda_1$ and $\lambda_2$, for a particular $\alpha$. If the spectrometer is set at the minimum $\Psi_{min}$, with the corresponding $\alpha$ where $\alpha=\beta$ for $\lambda_1$, then $n(\lambda_1)$ can be calculated from $\Psi_{min}(\lambda_1)$.
I was able to show ,[back in college many years ago=I was the T.A. (teaching assistant) for a course that did a laboratory experiment using the prism spectrometer along with the formula for the index n ], with a somewhat detailed calculation, that if you use the $\Psi(\lambda_1)$ above to compute $n( \lambda_1)$, you can also, to a very good approximation, find $n(\lambda_2)$, using $\Psi_{min}(\lambda_2) \approx \Psi_{min}(\lambda_1)+\Delta \beta$, where $\Delta \beta=\beta(\lambda_2)-\beta(\lambda_1)$, with $\alpha$ fixed at $\alpha=\beta(\lambda_1)$ as above. It turns out that $\Psi(\lambda_2)$ is also very near $\Psi_{min}(\lambda_2)$, so that $\Delta \Psi_{min} \approx \Delta \beta$, for $\lambda_1$ and $\lambda_2$. It's a lot of detail, but you might find it of interest. $\\$ In using a prism spectrometer, it is of interest to be able to compute (and measure) $\Delta \beta$ for two different wavelengths, from the graph of $n(\lambda)$ vs. $\lambda$. This is possible, with the approximation $\Delta \Psi_{min} \approx \Delta \beta$, along with the formula they give you for $n$ as a function of $\Psi_{min}$. $\\$ Note: To reach $\Psi_{min}(\lambda_2)$, it is necessary to change $\alpha$ to a new location, so that once again $\alpha=\beta$, but this time for $\lambda_2$. What essentially happens is the prism is rotated by an amount $\Delta \theta=\Delta \beta /2$. In the process, $\alpha$ increases, and $\beta$ decreases by that amount, with $\Psi$ remaining virtually unchanged, since near the minimum $\frac{d \Psi}{d \alpha} \approx 0$. This may be a lot of detail, but you might find it of interest. The spectrum tends to function as a unit when the prism is rotated, with wavelengths that are near each other having $\Delta \Psi_{min}=\Delta \beta$, where the $\beta$ is for some fixed $\alpha$, e.g. the $\alpha$ that makes for $\Psi_{min}(\lambda_1)$. The $\Psi(\lambda_2)$ was already very nearly at $\Psi_{min}(\lambda_2)$, even before this slight $\Delta \theta$ rotation.