Problem 54 in halliday, resnik, walker 9th edition Dispersion in a window pane.
ID: 1982944 • Letter: P
Question
Problem 54 in halliday, resnik, walker 9th edition
Dispersion in a window pane. a beam of white light is incident at angle =
50o on a common window pane (shown in cross section).
For the pane's type of glass, the index of refraction
for visible light ranges from 1.524 at the
blue end of the spectrum to 1.509 at the red end.
The two sides of the pane are parallel. What is the
angular spread of the colors in the beam (a) when
the light enters the pane and (b) when it emerges
from the opposite side? Hint: When you look at an
object through a window pane, are the colors in the light from the
object dispersed as shown in, say, Fig. 33-20?)
Explanation / Answer
Given that range of refractive index is from 1.524 to 1.509 Angle of incidence, 1 = 50 degree For the red light we have : From snell's law of refraction we have nairsin50 = n sin1 From the above we have sin50 = (1.509) sin1 1 = 30.5 degreesin50 = (1.509) sin1 1 = 30.5 degree
For the red light we have : From snell's law of refraction we have nairsin50 = n sin2 From the above we have sin50 = (1.524) sin2 2 = 30.2 degree For the red light we have : From snell's law of refraction we have nairsin50 = n sin2 From the above we have sin50 = (1.524) sin2 2 = 30.2 degree sin50 = (1.524) sin2 2 = 30.2 degree Then angular spread of colors is = 1 - 2 =30.5o - 30.2o = 0.3o
(b) If they emerges from the opposite side there is no difference in angle. So angular spread is zero.