Optics - Index of Refraction

A light source is at the bottom of a pool of water (the index of refraction of water is 1.33). At what minimum angle of incidence will a ray be totally reflected at the surface?

A. 0^o  
B. 25^o
C. 50^o 
D. 75^o
E. 90^o 

(GR9677 #56)

Solution:

Snell's Law: n_w sin θ_i = n_a sin θ_r

For total internal reflection, θ_r = 90^o
n_w sin θ_i  =  n_a
sin θ_i = n_a/n_w = 1/1.33 ≈ 3/4
Convert to degree, θ = ¾ × 180/π ≈ 45^o closer to 50^o 

Answer: C

Optics - Snell's Law

A beam of light has a small wavelength spread δλ about a central wavelength λ. The beam travels in vacuum until it enters a glass plate at an angle θ relative to the normal to the plate. The index of refraction of the glass is given by n(λ). The angular spread δθ' is given by

A. 

B. 

C.

D.

E.

(GR0177 #97)

Solution:

Snell's Law: n₁ sin θ₁ = n₂ sin θ₂

Given:
n₁ = 1 for vacuum
n₂ = n(λ) 
θ₁ = θ
θ₂  = θ'

sin θ = n(λ) sin θ'

Take the derivative of the equation with respect to λ:

^dsin θ/_dλ = ^{dn(λ)sin θ'}/_dλ ...(Eq.1)

θ is a constant → ^dsin θ/_dλ = 0

Eq.1 →
0 = ^{dn(λ)sin θ'}/_dλ 
0 = n(λ) (^dsin θ'/_dλ) + sin θ' (^dn(λ)/_dλ)   ...(Eq.2)

Since n is a function of λ, θ' is also a function λ →  
^dsin θ'/_dλ = (^dsin θ'/_dθ')(^dθ'/_dλ) = cos θ' (^dθ'/_dλ) 

Eq.2 →
0 = n(λ) cos θ' (^dθ'/_dλ) + sin θ' (^dn(λ)/_dλ) 
n(λ) cos θ' (^dθ'/_dλ) = − sin θ' (^dn(λ)/_dλ) 
n(λ) (^dθ'/_dλ) = − tan θ' (^dn(λ)/_dλ) 
^dθ'/_dλ = − (^tan θ'/_n(λ)) (^dn(λ)/_dλ) 
δθ' = |(^tan θ'/_n(λ)) (^dn(λ)/_dλ)|

Answer: E