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?
Best Answer
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...