Question

A ray of light falls on a denser rarer boundary from denser side .the critical angle is 45 degree. the maximum deviation the ray can undergo is

Answer 1

Given:

A ray of light falls on a denser rarer boundary from denser side .the critical angle is 45 degree.

To find:

Maximum possible deviation.

Calculation:

We have to consider the phenomenon of TOTAL INTERNAL REFLECTION , when light rays travel from denser to rarer medium $(\mu2>\mu1)$.

Since , 45° is the critical angle of incidence , if we slightly increase the angle of incidence , it will result in TIR.

Let's see the diagram

$\boxed{\setlength{\unitlength}{1cm}\begin{picture}(6,6)\put(1,3){\line(1,0){4}}\put(3,1){\line(0,1){4}}\put(3,3){\line(-1,-1){1}}\put(3,3){\vector(1,-1){1}}\put(3,3){\line(1,1){1}}\put(1,1){ \mu2 }\put(1,4){ \mu1 }\put(2,2){\vector(1,1){0.25}}\put(3.5,3.15){ \angle\delta }\end{picture}}$

So, the max deviation is coming as :

$\delta = {180}^{ \circ} - ( {45}^{ \circ} + {45}^{ \circ} )$

$= > \delta = {180}^{ \circ} - ( {90}^{ \circ} )$

$= > \delta = {90}^{ \circ}$

So, maximum deviation can be 90° , when the critical angle is 45° for a pair of media.