In the figure below, a light is incident at angle 1 = 40° on a series of five tr
ID: 1698563 • Letter: I
Question
In the figure below, a light is incident at angle 1 = 40° on a series of five transparent layers with parallel boundaries. For layers 1 and 3, L1 = 15 µm, L3 = 26 µm, n1 = 1.85, and n3 = 1.50.
(a) At what angle does the light emerge back into air at the right?
(b) How long does the light take to travel through layer 3?
In the figure below, a light is incident at angle theta 1 = 40 degree on a series of five transparent layers with parallel boundaries. For layers 1 and 3, L1 = 15 m, L3 = 26 mu m, n1 = 1.85, and n3 = 1.50. (a) At what angle does the light emerge back into air at the right? (b) How long does the light take to travel through layer 3?Explanation / Answer
(a) The exiting angle is 1 = 40°,
the same as the incident angle,
due to what one might call the “transitive”
nature of Snell’s law:
n1 sin 1 = n2 sin 2 = n3 sin 3 = …
from this equation we find out the angles (2 ,3, ....).
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(b)
Due to the fact that the speed (in a certain medium) is c/n .
where n = medium’s index of refraction
now we find
t = nL/c = (1.5)(26 × 10^-6 m)/(3.0 × 10^8 m/s)
t=13*10^-14 s =0.13 ps