Question

What is the angle of critical refraction if the upper layer has a velocity of 1400 m/s and the lower layer has a velocity of 2700 m/s?

Answer 1

I am assumimg that by 'upper layer has a velocity of 1400 m/s ' you are infact saying light in the upper medium travels at 1400 m/s .

From Snell's Law ;

(n_u)*(sin(i)) = (n_l)*(sin(r)) where 'n' is the refractive index of each medium , 'i' is the angle of incidence ,

'r' is the angle of refraction.

At critical refraction angle , r=90 degree =>sin(r)=1

=>sin(i)= (nl)/(n_u) => i= sin^-1 ((nl)/(n_u)).

But we know that n= c/v , c= speed of light in vaccum , v= speed of light in the medium

=>(nl)/(n_u) = v_u / v_l = 1400/2700= 14/27

=>i=sin^-1 (14/27) =31.24 degree