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Trigonometry / Optics problem
Nerissa
Registered User regular
Registered User regular
I'm in the process or refreshing my mathematical knowledge, and, to that end, I'm currently working through "Trigonometry DeMystified".
There's one problem in the book that I can't seem to follow why they did what they did. If anyone can explain it to me (or confirm my suspicions that it was an error in the book), I'd greatly appreciate it.
Problem:
Suppose a ray of white light, shining horizontally, enters a prism whose cross-section is an equilateral triangle and whose base is horizontal.
If the index of refraction of the prism glass is 1.52000 for red light and 1.53000 for blue light, what is the angle (delta) between rays of red and blue light as they emerge from the prism? Assume the index of refraction of the air is 1.00000 for light of all colors.
Their solution:
Find the angle of the red and the angle of the blue, then subtract the two.
For the red (which I am following fine):
First find the angle (rho1) of the red inside the prism
sin (rho1) / sin (30 deg) = 1.00000 / 1.520000
rho1 = 19.2049 deg
since the normal line to the surface is 30 degrees from horizontal, we subtract rho1 from 30 degrees to get the angle from horizontal, then add 30 degrees to get the angle to the other surface so:
rho2 = 30 - rho1 + 30 = 40.7951 degrees
now, find rho3
sin(rho3) / sin (rho2) = 1.00000 / 1.52000
rho3 = 83.2659 degrees
For the blue, I would assume you would follow the same steps, using 1.53000 in place of 1.52000.
sin(beta1) / sin(30 deg) = 1.00000 / 1.53000
beta1 = 19.0745 degrees
beta2 = 30 - beta1 + 30 = 40.9255 degrees
but... in finding beta3, they for some reason use 1.52 rather than 1.53?
sin(beta3) / sin(beta2) = 1.52000/1.00000
beta3 = 84.6952 degrees.
Oddly, if I use 1.53, I end up with sin(beta3) > 1, which is of course impossible... but what's the rationale for using 1.52 instead?
There's one problem in the book that I can't seem to follow why they did what they did. If anyone can explain it to me (or confirm my suspicions that it was an error in the book), I'd greatly appreciate it.
Problem:
Suppose a ray of white light, shining horizontally, enters a prism whose cross-section is an equilateral triangle and whose base is horizontal.
If the index of refraction of the prism glass is 1.52000 for red light and 1.53000 for blue light, what is the angle (delta) between rays of red and blue light as they emerge from the prism? Assume the index of refraction of the air is 1.00000 for light of all colors.
Their solution:
Find the angle of the red and the angle of the blue, then subtract the two.
For the red (which I am following fine):
First find the angle (rho1) of the red inside the prism
sin (rho1) / sin (30 deg) = 1.00000 / 1.520000
rho1 = 19.2049 deg
since the normal line to the surface is 30 degrees from horizontal, we subtract rho1 from 30 degrees to get the angle from horizontal, then add 30 degrees to get the angle to the other surface so:
rho2 = 30 - rho1 + 30 = 40.7951 degrees
now, find rho3
sin(rho3) / sin (rho2) = 1.00000 / 1.52000
rho3 = 83.2659 degrees
For the blue, I would assume you would follow the same steps, using 1.53000 in place of 1.52000.
sin(beta1) / sin(30 deg) = 1.00000 / 1.53000
beta1 = 19.0745 degrees
beta2 = 30 - beta1 + 30 = 40.9255 degrees
but... in finding beta3, they for some reason use 1.52 rather than 1.53?
sin(beta3) / sin(beta2) = 1.52000/1.00000
beta3 = 84.6952 degrees.
Oddly, if I use 1.53, I end up with sin(beta3) > 1, which is of course impossible... but what's the rationale for using 1.52 instead?
Nerissa on
0
Posts
The blue light will actually bounce back internally, instead of going out of the prism!
This explains critical angles with pretty diagrams.
I'm guessing it was a matter of copy/paste and forgetting to change that particular value, thus not discovering that the problem was set up badly.
I never thought to check the critical angle (obviously they didn't either), I guess because I assumed that the problem was set up in a way that made sense, given that it was designed explicitly to illustrate these concepts. (Also because they used the result of this problem in the next one.)
Anyway, so long as I have a basic grasp of the concepts and followed what they did, all is good. It's not like I'm explicitly studying optics at the moment (although it's now on my list), it was just a single chapter in the "applications" section of the book. One more chapter and then I go back to calculus.