Question

A ray of light hits a glass pane with 60o angle of incidence. The pane is 5 mm thick, with index of refraction 1.54. Light is reflected by a mirror that is in contact with the back of the pane. By how much is the beam laterally displaced with respect to the return path if the pane is removed?

Answer 1

As per the question,

Angle of incidence of the light beam$(i)=60^\circ$

Thickness of the pane $(t)=5mm =5\times 10^{-3}m$

refractive index of the pane $(n)=1.54$

Let the angle of refraction is =r,

$\sin (i)/\sin(r)=\mu$

$\Rightarrow \sin (r)=\dfrac{\sin(60^\circ)}{1.54}=\dfrac{\sqrt(3)}{2\times 1.54}=34^\circ$

Now, lateral displacement of the beam of light $=\dfrac{t}{\cos(r)}\sin(i-r)$

$=\dfrac{5\times 10^{-3}\sin(60^\circ-34^\circ)}{\cos(34^\circ)}=2.6mm$