Question

A light ray of wavelength 589 nm is incident at an angle θ on the top surface of a block of crown glass surrounded by air, as shown in the figure below. (The index of refraction of crown glass is 1.52.)

(a) Find the maximum value of θ for which the refracted ray will undergo total internal reflection at the left vertical face of the block.
°

(b) Repeat the calculation for the case in which the crown glass block is immersed in ethyl alcohol. (The index of refraction of ethyl alcohol is 1.361.)
°

(c) What happens if the block is immersed in carbon disulfide? (The index of refraction of carbon disulfide is 1.63.)

Answer 1

Angle of referation at top surface ' r ' is given by Snell's law
n1 sin i = n2 sin r
n1 and n2 are refrectice index of first and second medium where as i and r are incident and refrected angles.
here i =  θ , n1 = 1 and n2 = 1.52
Hence sin r = sin θ / n2
sin r = sin θ / 1.52

Angle of incident on left surface has to be greater than critical angle( C) for ray to undergo total internal reflecton.

C = sin^-1 (n2/n1) here ray is inside glass, so n1 = 1.52 and n2 = 1
C = sin^-1(1/1.52) = 41.14
r is also equal to 90 - C = 48.86....
sin r = 0.75 ......2
from 1 and 2 we have
sinθ = 0.75 x 1.52 >1
This tell us that for all incident angle θ , ray will under go referatin at left surface. That is even if incident at θ = 90 deg, angle of incident on left surface is less than critical angle.

b) When surrounded by medum of refrectice index 1.361
1.361 sinθ = 1.52 sin r

and r = 90 - C = 90 - sin^-1 (1.361/1.52) = 26.44
sin r = 0.44
sin θ = 1.52 x 0.44 / 1.361 = 0.497
θ = 29.8 deg
hence maximum angle of incidence is 29.8 deg for which ray undergoes total internal reflection.

c) Here refrective index of medium is more than glass, ray will enter the glass only if θ is less than critical angle for these two mediums that is sin^-1 (1.52/1.63). If entered in the glass if will always come out from left surface(provided it reaches left surface before reaching bottom surface).