Department of Energy Argonne National Laboratory Office of Science NEWTON's Homepage NEWTON's Homepage
NEWTON, Ask A Scientist!
NEWTON Home Page NEWTON Teachers Visit Our Archives Ask A Question How To Ask A Question Question of the Week Our Expert Scientists Volunteer at NEWTON! Frequently Asked Questions Referencing NEWTON About NEWTON About Ask A Scientist Education At Argonne UV vs X-ray
Name: sysop
Status: N/a
Age: N/A
Location: N/A
Country: N/A
Date: Around 1993


Question:
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?).



Replies:
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.

Arthur Smith


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.

Robert Topper



Click here to return to the Physics Archives

NEWTON is an electronic community for Science, Math, and Computer Science K-12 Educators, sponsored and operated by Argonne National Laboratory's Educational Programs, Andrew Skipor, Ph.D., Head of Educational Programs.

For assistance with NEWTON contact a System Operator (help@newton.dep.anl.gov), or at Argonne's Educational Programs

NEWTON AND ASK A SCIENTIST
Educational Programs
Building 360
9700 S. Cass Ave.
Argonne, Illinois
60439-4845, USA
Update: June 2012
Weclome To Newton

Argonne National Laboratory