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I have the following problem:

A thin beam of white light with negligible radius is incident at $50.0^\circ$ on a $10.0 cm$ thick slab of clear plastic. The index of refraction for red light in this material is $1.51$ and for violet light it’s $1.54$. Determine the approximate diameter, $D$ of the emerging beam. Express your answer in cm with two significant figures.

I realise that this is a physics problem, and asking physics questions is off-topic on math.stackexchange. However, my question does not pertain to the physics of the problem, but rather the geometry used to solve certain aspects of it.

This seems to be a variation (same problem, different numeric values) of problem 4.20 in Optics, 4th edition, by Hecht.

My solution proceeds as follows:

Snell's law can be used to find $\theta_{red}$ and $\theta_{violet}$:

$$\begin{align} \theta_{red} &= \arcsin \left[ \dfrac{n_1}{n_{red}} \sin(\theta_i) \right] \\ &= \arcsin \left[ \dfrac{1}{1.51} \sin(50^\circ) \right] \end{align}$$

$$\begin{align} \theta_{violet} = \arcsin \left[ \dfrac{1}{1.54} \sin(50^\circ) \right] \end{align}$$

Since $\tan(\theta) = opposite/adjacent$, we have that

$$\begin{align} h_{red} = 10 \tan(\theta_{red}) \end{align}$$

$$\begin{align} h_{violet} = 10 \tan(\theta_{violet}) \end{align}$$

Now, here is where I ran into trouble. I now need to find the diameter of the emerging beam. My instructor's solutions give the diameter of an emerging beam as $D = \cos(50^\circ)(h_{red} - h_{violet}$).

The following is an illustration of the problem in question:

enter image description here

However, when researching how to find the diameter of an emerging beam, I found conflicting answers. The solutions to the aforementioned Hecht textbook give the diameter of the emerging beam as $D = \dfrac{h_{red} - h_{violet}}{\cos(50^\circ)}$. However, I have also found some documents that use the same formula as my instructor. Furthermore, I found two YouTube videos that claim to derive the formula for the diameter of an emerging beam, but derive different answers!

After spending much time researching this, it seems that there is no clear consensus, at least, none that I can find, on what the formula for the diameter of an emerging beam actually is. And another problem is that, for a learner such as myself who wants to actually understand the formula (rather than blindly memorizing some formula that was handed to me), there are no sources that provide a clear, step-by-step derivation this formula. At this point, I am completely lost.

I am kindly requesting that someone knowledgeable on this subject please take the time to (1) provide the formula for the diameter of an emerging beam, and (2) provide a step-by-step, well explained proof/derivation of that formula.

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Let's call $V$ the point where the violet beam exits the plastic, and $R$ where the red beam exits the plastic. Draw a perpendicular from $R$ to the outgoing violet beam, and call $P$ the point of intersection. In this right angle triangle, the hypotenuse is $h_{red}-h_{violet}$ and one of the sides is $D$. Notice that all outgoing beams will exit the plastic at the same angle with respect to the normal, so at $50^\circ$ with respect to the normal to the surface. Then the angle between the beam and the direction parallel to the surface is $40^\circ$. So in the $\triangle VRP$ you can write $$\sin 40^\circ=\frac{D}{h_{red}-h_{violet}}$$ Note that $\sin 40^\circ=\cos 50^\circ$, so $$D=(h_{red}-h_{violet})\cos50^\circ$$

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  • $\begingroup$ "Notice that all outgoing beams will exit the plastic at the same angle with respect to the normal"; Why is this the case? Please justify this. $\endgroup$ – The Pointer Nov 4 '19 at 6:37
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    $\begingroup$ You have parallel surfaces, and the same media on both sides. You start with $\theta_i$ on the first interface and the refracted beam makes an angle $\theta_r$ with the normal. Then the incident beam on the second surface will be at the same $\theta_r$, and, using the index of refraction, the outgoing beam will make the same $\theta_i$ angle with the normal $\endgroup$ – Andrei Nov 4 '19 at 7:01
  • $\begingroup$ Ok, thanks for the clarification. So, in a matter of speaking, the index of refraction "undoes" the changes it made to the beam angle when it entered the media, now that it is leaving the media? $\endgroup$ – The Pointer Nov 4 '19 at 9:08
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    $\begingroup$ Yes, as long as you have parallel faces and the same media on both sides $\endgroup$ – Andrei Nov 4 '19 at 12:47

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