Question

Two plane parallel plates having thickness of 3cm and 5cm and refractive indices of 1.5 and 1.8 respectively are piled together .A light beam incident on the first plate at an angle of 60' to the normal ,passes through the plates and emerges at the same angle .Find the displacement of the beam.

Answer 1

Answer on Question 38092, Physics, Optics From the picture in attachment one can easily find that

$h=AB\sin(\alpha-\phi)=\frac{d}{\cos\alpha}\sin(\alpha-\phi)$

We will use this formula for displacement of beam with parallel plate

$\Delta x=\frac{d}{\cos\alpha}\sin(\alpha-\beta)$

where $\alpha$ is angle of incidence, $\beta$ is angle of refraction and d is thikness. We can find those angles for both plates easily from Snell law:

$\beta_{1}=\arcsin\frac{\sin\alpha}{n_{1}},\qquad\beta_{2}=\arcsin\frac{\sin\alpha}{n_{2}}$

So we find

$\Delta x=\Delta x_{1}+\Delta x_{2}=\frac{d_{1}}{\cos\alpha}\sin(\alpha-\beta_{1})+\frac{d_{2}}{\cos\alpha}\sin(\alpha-\beta_{2})=$

$=\frac{d_{1}}{\cos\alpha}\sin(\alpha-\arcsin\frac{\sin\alpha}{n_{1}})+\frac{d_{2}}{\cos\alpha}\sin(\alpha-\arcsin\frac{\sin\alpha}{n_{2}}))\approx 7.7\,cm$