Index Key:  PHY081
Author:     sysop
Subject:    UV vs X-ray
Text:       Why does violet refract more through a prism than red when its 
frequency is higher and hence it may have more penetrating ability (i.e. UV 
vs. X-ray?).

Response #:  1 of 2
Author:      Arthur Smith
Text:        If I recall correctly, light refraction and absorption are 
connected, in that if you have a frequency of light that is strongly absorbed, 
then the amount of refraction is strongest on either side of that absorption 
frequency, and gets weaker as you go further away.  I think that in a prism, 
for example, the absorption frequencies are at quite high energies (where the 
electrons start to get excited), maybe in the high UV.  The glass is 
transparent because it has a "band gap" of maybe 1 electron- volt, and so that 
would be where the first absorption occurs.  Anyway, that means that, at 
frequencies well below this absorption frequency, such as at visible light 
frequencies, the refraction increases with increasing frequency.  At very high 
frequencies (such as X-rays) the refraction decreases again, because you are 
on the other side of the absorption peak.  Right in the absorption region, of 
course, light is not refracted, but is absorbed by the material, and it 
becomes opaque.



Response #:  2 of 2
Author:      Robert Topper
Text:        That does sound reasonable.  Perhaps we can sum it up by saying 
that the refractive index of the prism has a strong wavelength dependence, 
which just happens to be in the visible part of the spectrum (for the reasons 
you have stated)?  Snell's law says that light will bend as it enters the 
prism and as it exits on the opposite face, and that the angle of each "bend" 
is given by n1 sin(theta1) = n2 sin(theta2).  Here n_1,n_2 are (respectively) 
the index of the air and the glass when considering the first reflection, and 
the index of the glass and the air on the second reflection.  BUT..this 
equation is only valid for a single wavelength of light.  If the refractive 
index of the glass has a strong wavelength dependence, then the exit angles 
will be different for each wavelength.  (Assuming that the refractive index of 
air does not depend much on the wavelength).  Now, the student could actually 
figure out what the wavelength dependence was by measuring the entrance and 
exit angles of the various wavelengths and using Snell's law...a fun 
experiment.  Just a little expansion on Arthur Smith's (correct) answer.