Homework #9 (chapter 39): Light Waves

Exercise 9:

t = d/c è Dt = Dd/c = 2*10^-2 m / 3*10⁸ m/s = 6.67*10^-11 s

Exercise 19:

With q_i being the angle of incidence (between incident ray and left vertical container wall) and q_r the angle of refraction (between refracted ray and normal to surface)

n_rsinq_r = n_isinq_i = 1 since n_r»1 and q_r=90°

==> n_i = 1/sinq_i = 1/[1.14m/(1.14²m²+0.85²m²)] = 1.25

Exercise 40:

For u<<c, f’ = f (1-u/c)

c = f * l = f’ * l’ ==> l’ = fl/f’ = c/f’ = c/[f(1-u/c)]

Then Dl/l = (l’-l)/l = l’/l - 1 = {c/[f(1-u/c)]}/(c/f) – 1 = (u/c)/(1-u/c) » u/c for u/c<<1

Problem 1: will be discussed in class

Problem 9:

a) Without the liquid on the slab direct a light beam to the center. Vary q_i until q_i = q_c (critical angle of total internal reflection). Then q_i = q_c = arcsin(n_air/n_glass) = arcsin(1/n_glass) è n_glass = 1/sinq_c

Add the liquid and find again the angle of total internal reflection, now q_c’

q_i’ = q_c’ = arcsin(n_liquid/n_glass) è n_liquid = n_glass sinq_c’

b) Not very practical, since n_liquid < n_glass is a restriction to the method. Furthermore, the slab can not be tilted or the liquid will run off, i.e. the beam has to be moved maintaning the alignment which is difficult to achieve.