(59th Polish Olympiad in Physics, final stage, experimental part, 2010)

You have at your disposal:

  • a sample of blue foil of a homogeneous material, placed between two glass panes in a slide frame
  • a laser pointer
  • a meter of the power of light composed from a photodiode, a battery and a voltmeter whose readings are proportional to the power of light falling on the active surface of the photodiode
  • graph paper
  • two wooden blocks
  • adhesive tape

Determine the refractive index of the material the blue foil is made of for the laser wavelength of light.

Note 1: The thickness of the foil is approximately 0.1 mm and the thickness of the glass panes on its both sides is approximately 1 mm. Between the panes and foil there is a very thin layer of a liquid, which has a refractive index very close to the refractive index of glass.

Note 2: The frame is not shut completely - it should not be pressed or opened. Make sure not to smear the surface of the panes.

There is no official solution available for this problem. How could it be solved?

  • $\begingroup$ While this is not homework in the strict sense, it is homework-like in terms of our working definition. I think it is a fun question, and would love to see it survive (though that may require this to be rephrased as a conceptual problem, instead of this straight-from-the-questionnaire form) $\endgroup$
    – 299792458
    Apr 8, 2015 at 18:35

2 Answers 2


I don't know the "official" answer but here is what I might try. I am hoping that others will contribute to make this a "good" answer.

First - we were not told whether the wavelength of the laser is transmitted at all by the blue foil; but since blue foil typically absorbs red light, and most laser pointers are red (I have a blue one but they are expensive!) I will assume we have no transmission.

That means we need to determine the answer with reflection. The Brewster angle may come to our rescue here. Since a laser beam is polarized, there is a certain angle for which we see no reflection from a surface when the polarization is in the plane containing the normal to the surface and the path of the beam. It should be fairly easy to set up the laser pointer at the Bragg angle (just look at the reflected spot and play around with both the angle of incidence, and the rotation of the laser pointer). Use the ruler to determine the angle (I assume you are allowed a calculator for this exercise - or trig tables, or a good slide rule. This was not specified. If not, then some origami on the graph paper will get you a pretty good goniometer...)

This will give us one data point: since the Brewster angle $\theta_B$ at the interface of materials with refractive index $n_1$ going to $n_2$ is given by

$$\theta_B = \tan^{-1}\left(\frac{n_2}{n_1}\right)$$

the refractive index $n_2$ of the slide glass is given by

$$n_g = \tan\theta_B$$

Now comes the tricky part: we want to try to find the point of partial extinction of the second reflection - the one off the interface between the glass and the foil.

enter image description here

For this we want to carefully plot the intensity of the laser beam as a function of angle; I suspect we are looking for a secondary dip in the curve where the reflection from the back surface of the glass / foil interface is completely gone. This will happen when the internal angle ($\theta_2$ in my diagram) obeys the Brewster angle relationship.

We rewrite that relationship as

$$n_{foil} = n_g \tan\theta_2$$

and we know from Snell's Law that


It is possible that there is a range of refractive index values for which there is no solution. Specifically, the angle $\theta_2$ is limited by Snell's Law to be less than $\sin^{-1}\frac{1}{n_g}$, so this approach will not work if

$$\sin\tan^{-1}\frac{n_{foil}}{n_g} \lt \frac{1}{n_g}$$

This answer is not guaranteed to be error free... I need to eat something and revisit this (hope there will be some constructive comments / edits in the meantime).


After reading Chris Mueller's answer, I decided to have another go and plot these curves for the reflected power given two different refractive indices, $n_1$ and $n_2$ (where there is a further $n_0=1$ for air).

Using the Fresnel equations we can compute the sum of reflected intensity for both the "front" and "back" reflections - assuming that the second reflection will in turn be partially reflected at the glass-air interface. There will be multiple internal reflections - each a small fraction of the previous one. There may also be reflections off the back of the foil; but as I said before I am going to assume that red laser pointer light is completely absorbed by the blue foil. The whole system now looks like this:

enter image description here

Then we have to use the following expressions a number of times (suffixes $s$ and $p$ represent polarization - either parallel to the surface, or perpendicular) - I am putting the incident intensity $i_0=1$ for simplicity (note - updated to show coefficients for intensity not amplitude with thanks to Rob Jeffries for pointing out my mistake):

$$R_s=\left|\frac{n_0 \cos\theta_0-n_1\cos\theta_1}{n_0 \cos\theta_0+n_1\cos\theta_1}\right|^2\\ T_s = 1 - R_s$$ $$R_p=\left|\frac{n_0 \cos\theta_1-n_1\cos\theta_0}{n_0 \cos\theta_1+n_2\cos\theta_0}\right|^2\\ T_p = 1 - R_p$$

Now the total reflected intensity is an infinite sum: if we put the coefficient of reflection at the first interface as $R_1$, the second as $R_2$, and the reflection on the inside of the glass-air interface as $R_3$, then we can write for the corresponding transmissions $T_1 = 1 - R_1, T_3 = 1 - R_3$ (we are assuming everything that is transmitted into the foil is absorbed). Then we have for the total reflected intensity:

$$\begin{align}R_t &= R_1 + (T_1 R_2) T_3 + T_1 R_2 R_3 R_2 T_3 + ...\\ &=R_1 + T_1 R_2 T_3 \left(1 + R_2 R_3 + (R_2 R_3)^2 + ...\right)\\ &= R_1 + \frac{T_1 R_2 T_3}{1 - R_2 R_3}\\ \end{align}$$

This gets pretty messy pretty quickly - but that's why we all carry little supercomputers in our pockets these days...

I am going to assume that the laser is polarized and aligned per the above, so we can use just the expression for $R_s$; then something very interesting happens. At the Brewster angle, we will see a minimum in the reflectance - but it will not be zero, since the transmitted beam will undergo partial reflection at the interface between glass and foil. So while $R_1$ in the above expression would be 0, the other terms are not. In fact, it's quite easy to evaluate the expression for a range of refractive index values: and when you do this, you get the following plot:

enter image description here

The Python code I used to generate this plot (note - this does not take account of the correction for the square-of-the-amplitude pointed out above so it will give the wrong quantitative result although it shows the null at the right place):

# use Fresnel to compute reflection from composite surface
import numpy as np
import matplotlib.pyplot as plt
import math

def fresnel(theta1, n1, n2):
    n12 = n1 / n2
    st = n1/n2 * math.sin(theta1)
    if st>1:
        return(1, 0)
        theta2 = math.asin(st)
        r = np.abs( (math.cos(theta1)-math.cos(theta2)*n12)/(math.cos(theta1)+math.cos(theta2)*n12))
        return (r, theta2)

theta = np.arange(0, math.pi/2, 0.01)
n2 = np.arange(1, 3, 0.01)
theta0 = math.atan(1.52) # start at Brewster angle
(r1, theta1) = fresnel(theta0, 1.0, 1.52)
t1 = 1 - r1
# second reflection: range of values of n
rt2 = np.array([fresnel(theta1, 1.52, n) for n in n2])
r2 = rt2[:,0]

# probability of light escaping:
(r3, theta3) = fresnel(theta1, 1.52, 1.0)
t3 = 1 - r3

# combine the geometric series:
power = r1 + (t1 * r2 * t3) / ( 1 - r2 * r3)

# plot the result:

plt.plot(n2, power)
plt.xlabel('foil index')
plt.ylabel('reflected power')
plt.title('compound reflection')

Now if the refractive index of the foil is greater than the refractive index of glass, this gives us an easy way to determine the value by looking at the reflected power: over the range of values I considered, it is almost a straight line.

However, if $n_2 < n_1$, we have that funny curve to the left of the Brewster dip, which is not very helpful. I am still puzzling over how you would deal with that case.

  • 1
    $\begingroup$ Crappy blue lasers are no longer expensive: amzn.com/B00J7IFESY $\endgroup$ Apr 8, 2015 at 21:34
  • $\begingroup$ @romkyns be REALLY careful. That laser should probably be illegal. I think that is a 405 nm laser - it is cheap because it is used in DVD players. It is very powerful ("<50 mW") but almost invisible. Until you hit a piece of white paper which fluoresces BRIGHTLY as the power moves to "visible" part of the spectrum. Play with these for a long time without glasses and you may go blind because of the UV. I had one and after a few minutes I realized I had developed a "blind spot" near the center of my field of view. It took a while to recover... That is NOT a blue laser. 450 - 495 nm is blue. $\endgroup$
    – Floris
    Apr 8, 2015 at 22:35
  • $\begingroup$ Thanks for the warning. Mine says 405nm on it, and was actually sold as a UV laser, <10 mW. It looks violet when it hits most surfaces, but paper fluoresces rather brightly in a different shade. I do rather hope they don't become illegal. Motorcycles are vastly more dangerous and they're legal! $\endgroup$ Apr 8, 2015 at 22:45
  • $\begingroup$ @romkyns - the problem is that these things are dangerous and you don't know it. I hope yours is less than 10 mW - but the Amazon link you states (in response to a question "how is the beam power?"): "The tag on the laser says the power is less than 50 mw, 405 nm +- 10 nm" . $108 buys a "real blue" laser pointer: z-bolt.com/blue-laser-pointers/blp-5.html . I am not associated with that company... $\endgroup$
    – Floris
    Apr 8, 2015 at 22:49
  • $\begingroup$ I think you have the definition of the intensity reflection coefficients wrong. These are expressions for amplitudes. This wouldn't matter, except where you have used $T_p = 1 -R_p$, which is not true for the amplitudes. You need to square your reflection coefficients first. $\endgroup$
    – ProfRob
    Aug 24, 2015 at 13:50

Use your power meter and graph paper to map out the reflectivity as a function of angle. The Fresnel equations (plots shown below) are rather sensitive to the relative indices of refraction of the glass-foil interface. If the index of refraction of the foil is higher than the glass then you will hit a plateau above the critical angle at which all of the power will be reflected. If the index of refraction is smaller than the glass, then you should be able to identify the Brewster's angle dip in the reflectivity.

Some caveats:

  • You will need to be careful to seperate the other reflections because the air/glass interfaces will also obey the Fresnel equations. This should be relatively easy at high angles of incidence, but might be impossible at lower angles.
  • If the laser is polarized, then you will want to rotate it to minimize the reflectivity at each point which will ensure that you are using p polarization. If it is unpolarized (which is unlikely), then you will have to be very careful to identify the Brewster dip.

Fresnel equations plots

  • $\begingroup$ Can you comment on what the suffixes $s$ and $p$ in your graphs stand for? And how do these curves look different when $n_1$ and $n_2$ are both greater than 1 (as they will be)? I think you are on your way to having the right answer here... but you need to plot the equations for the combination of two different materials - and then figure out how to solve for $n_2$ (assuming there is a $n_0=1$ representing air). You have encouraged me to take another look and maybe add the full calculation to my answer. $\endgroup$
    – Floris
    Apr 8, 2015 at 22:52
  • $\begingroup$ @Floris $s$ and $p$ stand for the different polarization states. You are right that this is just a sketch of one way to do it; maybe I'll add to it tomorrow. The plots I posted are taken straight from the Wikipedia article that I linked, but they won't change significantly for a second medium with $n\neq 1$. $\endgroup$ Apr 9, 2015 at 1:12

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