Question

(a) Prove that a ray of light incident on the surface of a sheet of plate glass of thickness $$t$$ emerges from the opposite face parallel to its initial direction but displaced sideways,as in above figure. (b) Show that, for small angles of incidence $$\theta$$, this displacement is given by $$x=t\theta \frac{n-1}{n}$$.
where $$n$$ is the index of refraction of the glass and $$\theta$$ is measured in radians.

Answer 1

Let $$θ$$ be the angle of incidence and $$\theta_{2}$$ be the angle of refraction at the left face of the plate. Let $$n$$ be the index of refraction of the glass. Then, the law of refraction yields $$\sin \theta =n\sin \theta _{2}$$. The angle of incidence at the right face is also $$\theta _{2}$$. If $$\theta _{3}$$ is the angle of emergence there, then $$n\sin \theta_{2} =\sin \theta _{3}$$. Thus $$\sin \theta_{3} =\sin \theta$$ and $$\theta_{3}=\theta$$.
The emerging ray is parallel to the incident ray. We wish to derive an expression for $$x$$ in terms of $$θ$$. If $$D$$ is the length of the ray in the glass, then $$D\cos \theta _{2}=t$$ and $$D=t/\cos \theta _{2}$$. The angle $$α$$ in the diagram equals $$\theta - \theta _{2}$$ and
$$x=D\sin \alpha =D\sin \left ( \theta -\theta _{2} \right )$$,
$$x=\frac{t\sin \left ( \theta -\theta _{2} \right )}{\cos \theta _{2}}$$.
If all the angles $$θ$$, $$\theta _{2}$$, $$\theta _{3}$$ and $$\theta - \theta _{2}$$ are small and measured in radians, then $$\sin \theta \approx \theta$$, $$\sin \theta_{2} \approx \theta_{2}$$, $$\sin\left ( \theta -\theta _{2} \right ) \approx \theta -\theta_{2}$$ and $$\cos \theta _{2}\approx 1$$. Thus $$x\approx t\left ( \theta -\theta _{2} \right )$$. The law of refraction applied to the point of incidence at the left face of the plate is now $$\theta \approx n\theta _{2}$$, so $$\theta _{2}\approx \theta /n$$ and
$$x\approx t\left ( \theta -\frac{\theta}{n} \right )=\frac{\left ( n-1 \right )t\theta}{n}$$.