# Calculating light's lateral shift in a glass slab

I have tried to compute light's lateral shift after passing through a glass slab

In terms of the incidence angle $\theta_1$, the slab thickness $d$ and the refraction index $n$, I have found that the lateral shift $x$ is given by

$$x = \frac{d}{n^2} \sin(\theta_1) \left( \sqrt{n^2 + \sin^2(\theta_1)} \right) \left( \sqrt{n^2 - \sin^2(\theta_1)} - \cos(\theta_1) \right).$$

In order to test this formula I have considered two extreme cases: when the light is perpendicular to the slab ($\theta_1 = 0$) and when it is almost parallel ($\theta_1 \approx \pi/2$). The first case works OK, since if $\theta_1 = 0$ then $\sin(\theta_1) = 0$ and $x=0$. This is physically plausible, for if the light ray is perpendicular then it does not change direction.

I'm having trouble assessing the second case. If $\theta_1 \approx \pi/2$ then the formula can be written as

$$x = \frac{d}{n^2} \sqrt{n^4 -1}.$$

Is this physically plausible? My guess has been that if I can say that $n^4 \gg 1$ I obtain $x \approx d$. This sounds correct to me since if the light ray approaches the slab almost parallel then after refracting back to air it should come out almost parallel, causing a lateral shift of exactly $d$. I am not sure, however, in saying $n^4 \gg 1$. I have checked glass's refraction index and it lies in the range $1.5 - 1.9$. Am I correct?

• That's not actually the lateral shift, lateral would be parallel to the plane of the glass. Still it is a simple problem, see my answer below.
– DrM
Jan 8, 2022 at 23:56

I get a different formula. Let me show you how I derived it.

Using the following diagram:

We can write the following equations by looking at triangles:

\begin{align}\frac{x}{L} &= \sin(\theta_1-\theta_2)\\&=\sin\theta_1 \cos\theta_2 - \cos\theta_1\sin\theta_2\\ \frac{d}{L}&=\cos\theta_2\end{align}

Assuming that the air has a refractive index of 1, we can further write

$$\frac{\sin \theta_1}{\sin\theta_2}=n$$

From basic geometry we know that for angles in the first quadrant,

$$\cos\theta = \sqrt{1-\sin^2\theta}$$ Combining these gives

\begin{align}x &= \frac{d}{\cos\theta_2}\left(\sin\theta_1 \cos\theta_2 - \cos\theta_1 \sin\theta_2 \right)\\ &= d\left(\sin\theta_1 - \frac{\sin\theta_1\cos\theta_1}{n\cos\theta_2}\right)\\ &=d\sin\theta_1\left(1-\frac{\sqrt{1-\sin^2\theta_1}}{n\sqrt{1-\frac{\sin^2\theta_1}{n^2}}}\right)\\ &=d\sin\theta_1\left(1-\frac{\sqrt{1-\sin^2\theta_1}}{\sqrt{n^2-\sin^2\theta_1}}\right)\end{align}

Note that with this expression, the distance $x$ will approach $d$ when $\theta_1$ approaches $\pi/2$ since the second term will vanish.

You might want to compare my approach with yours. I'm not claiming mine is right...

• Neat derivation. Thank you. In my approach I computed $L$ and the substituted, instead of using it as a reference element. Aug 13, 2015 at 11:40
• The working checks out for me. It's quite curious that the $n\to1$ and $\theta_1\to\pi/2$ limits don't commute. Aug 13, 2015 at 11:46
• The discussion of the incidence angle close to 90 degrees is just a mathematical problem. Physically, due to total internal reflection, the light will not exit in the air on the other side. This will happen at angles that are not close to 90 degrees but around 40 degrees for glass. The slab will act like a wave guide after this value. So there is no physical meaning to the discussion of incident angles near 90 degrees.
– nasu
Sep 23, 2016 at 17:30
• @nasu I don't think that is true. If the light starts outside the slab with an angle less than 90 degrees, the light inside the slab will have an angle less than the critical angle and some light will come all the way through to the other side. I agree that $\theta_2$ cannot get close to 90 degrees - but I never claimed that. Sep 23, 2016 at 19:05
• $L=t\left(\frac{\sin\left(i-\sin^{-1}\left(\frac{\sin\left(i\right)}{\mu}\right)\right)}{\cos\left(\sin^{-1}\left(\frac{\sin\left(i\right)}{\mu}\right)\right)}\right)$
– sato
Jan 11, 2021 at 7:57

$$\left[\text{lateral shift}\right]~=~\left[\text{thickness}\right] \frac{\sin{\left(i-r\right)}}{\cos{\left(r\right)}} \,,$$where:

• $\left[\text{thickness}\right]$ is the thickness is of the glass slab;

• $i$ is the angle of incidence;

• $r$ is the angle of refraction.

• $r = sin^{-1}\left(\frac{sin(i)}{\mu}\right)$
– sato
Jan 11, 2021 at 7:48

That is indeed the $\theta\to\pi/2$ limit of the expression you give. This is physically reasonable because you require $x$ to be zero when $n=1$ as there is then no glass slab to deflect the beam.

For the region in between $n=1$ and $n\gg1$, there's not all that much that can tell you whether the result is reasonable or not. The change in $x$ from $0$ to $d$ as $n$ goes from $1$ to $n\gg 1$ is continuous but quick. If you graph it, you get a very sudden uptick at $n=1$,

$\hspace{125px}$,

and by the time it reaches $n\approx 1.5$, which is where your intuition about how glass behaves was formed, $x$ is already at 90% of $d$. Again, that's reasonably consistent with intuition. If you want more validation, though, you'll just need to double-check your calculations, or reproduce them inside a bigger framework which you trust more.

• @MarkFantini Yes. Source code in html comments in the answer source (click edit), with ->s removed. Aug 13, 2015 at 11:25

This case is not practically possible that lateral shift can equal the thickness of the slab. Because for that, the ray after refraction from the first surface must travel obliquely i.e. at an angle of $90^\circ$ with the surface of slab and after reaching the other surface it must bend $90^\circ$ and graze the surface...but this doesn't happen.

That's it. So we can say that we can have maximum lateral shift for $i\rightarrow 90^\circ$. This is just hypothetical and we can't prove this practically.

• Why is their limit not practical? As $\theta_{1} \rightarrow 90^{\circ}$ the asymptotic limit should come close to $d$, should it not? I do not think your statement answers their question. Mar 3, 2016 at 13:27

First of all, some define lateral displacement as the distance along the direction parallel to the surface and the quantity you refer to as the beam displacement. To me that makes most sense in terms of how it effects experiment design and analysis. But, in any case, both are easy to calculate. Here is a simple way to think about it.

By inspection, we see that the displacement along the surface, which we will call "x", is given by the tangent of the internal angle, times the thickness of the slab. Combining that with Fresnel's equation, we have

$$x = d \tan( \theta_{internal} ) = d\ \tan(\sin^{-1}({\sin(\theta)\over n}) )$$

Now for the distance you asked for, the displacement along a normal to the original beam.

Again by inspection, we see that the shift x along the surface is the hypotenuse to a triangle formed with the shift you asked for, as its base and with internal angle equal to the incident angle.

Lets call the shift you asked for as "u". Thus

$$u = x \cos(\theta) = d \cos(\theta) \tan( \sin^{-1}(\sin({\theta\over n}) )$$

That said, for the intensity of the displaced beam, you have to considered internal reflection. So, the intensity is going to vary a little bit with angle. If you need an accurate model, for example to correct some experimental data, then depending on the range of angles and wavelengths relative to the thickness of the slab, you might need to include that effect and integrate over wavelength and aperture.

• You might enjoy learning how to format your equations with Mathjax: see math.meta.stackexchange.com/q/5020/101979 Jan 11, 2022 at 5:50
• @Floris, very nice, I was not aware that exchange supported mathjax
– DrM
Jan 12, 2022 at 15:06