Question

A ray of light is incident in water to a plane water/air boundary at an angle of incidence of 45o. Calculate the angle of refraction of the refracted ray given the refractive index of water is 1.33

Calculate the critical angle for the water/air boundary

Answer 1

Answer:

angle of refraction =32.1, critical angle =48.8

Explanation:

The formula for finding refractive index(snell's law) is $\frac{sin(i)}{sin(r)}$, the formula can work vice versa. Hence , for solving this qn first we know the give refractive index is 1.33 and the angle of incidence is 45 degrees thus, $\frac{sin(45)}{sin(x)} =1.33$ i.e x is the value of the angle of refraction. We make x the subject and hence we will get $sin^{-1} (\frac{sin(45)}{1.33})=32.1 (3.s.f)$. This is the angle of refraction and this would answer the first part.  

The second  part on finding the critical angle is very simple as another formula for refrctive index is $\frac{1}{sin(c)} =1.33$ ( where c is the critical angle)

$c= sin^{-1} (\frac{1}{1.33} )=48.8(3.s.f)$. Thus the answer for the critcial angle would be 48.8 degrees.

Hope this helps :))