Question

A ray of light strikes a glass slab of thickness $t$
(a) Prove that the ray is deflected laterally by distance
$t\sin \theta [1-\sqrt{\frac{1-{\sin }^2\theta }{n^2-{\sin }^2\theta }}]$
(b) Prove that for a small angle of incident the lateral shift $y$ is given by $y=t\theta (1-\frac{1}{n})$
When $n$ is the refractive index of glass with respect to air

Answer 1

(a) Lateral shift, $y=BC=AB\sin r=AB\sin (\theta -r)$
$\Rightarrow y=\frac{t}{\cos r}(\sin \theta \cos r-\cos \theta \sin r)=t(\sin \theta -\cos \theta \tan r)$
$=t\sin \theta (1-\cot \theta \tan r)$ $\left(1\right)$
Where $AB=t/\cos r$ and $r=$ angle of refraction given by using Snells Law,
$\frac{\sin \theta }{\sin r}=n$
$\Rightarrow \sin r=\frac{\sin \theta }{n}$ & $\cos r=\sqrt{\frac{n^2-{\sin }^2\theta }{n}}$
Putting the values of $\sin r$ & $\cos r$ in $\left(1\right)$ we obtain
$y=t\sin \theta [1-\cot \theta \frac{\frac{\sin \theta }{n}}{\sqrt{\frac{n^2-{\sin }^2\theta }{n}}}]=t\sin \theta [1-\frac{\cos \theta }{\sqrt{n^2-{\sin }^2\theta }}]$
$=t\sin \theta [1-\sqrt{\frac{1-{\sin }^2\theta }{n^2-{\sin }^2\theta }}]$
(b)If $\theta$ is small, ${\sin }^2\theta$ can be ignored in comparison to $1$, $n^2$
$\Rightarrow y\cong t\theta [1-\sqrt{\frac{1}{n^2}}]$
$\Rightarrow y=t\theta [1-\frac{1}{n}]$.