Index of refraction and electromagnetic radiation

AI Thread Summary
The discussion revolves around calculating the index of refraction for electromagnetic radiation with a frequency of 5 x 10^14 Hz and a speed of 2.54 x 10^8 m/s. The formula for the index of refraction (n) is established as the ratio of the speed of light in a vacuum to the speed of light in the medium. Amy initially sought help but later figured out the solution herself. Additionally, she explored another problem involving apparent depth and angles of refraction, ultimately calculating the actual depth of an object underwater. The conversation highlights the application of refraction principles in different scenarios.
physicsgal
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"An electromagnetic radiation has a frequency of 5 x 10^14Hz

a) (wave in vacuum = 600nm)
b) (wave in water = 440 nm)
c) what is the index of refraction of a medium in which the speed of this radiation is 2.54 x 10^8 m/s?"

so given: c = 2.54 x 10^8 m/s
f = 5 x 10^14Hz

required: ni

im not sure where to start on this one. any help will be appreciated.

~Amy
 
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nevermind, i figured it out! :biggrin:

n = C/v

i don't know how to delete my original post.

~Amy
 
physicsgal said:
"An electromagnetic radiation has a frequency of 5 x 10^14Hz

a) (wave in vacuum = 600nm)
b) (wave in water = 440 nm)
c) what is the index of refraction of a medium in which the speed of this radiation is 2.54 x 10^8 m/s?"

so given: c = 2.54 x 10^8 m/s
f = 5 x 10^14Hz

required: ni

im not sure where to start on this one. any help will be appreciated.

~Amy

I can help with part c, the index of refraction is a ratio between the speed of light in a vacuum to the speed of light in the medium, so if we call the index of refraction n, then

n = (speed of light in vacuum)/(speed of light in medium)

For parts a and b, I think you can do a similar thing with a ratio of wavelengths, but I'm not sure if that's right or not so you may want to wait for someone to give you a more concrete answer about that.
 
thanks dLeet!

im working on another problem if anyone can share some insight..

Someone is on their boat and their eye level is 1 m above the water. and they guess the apparent depth of an object is 2 m below the water surface. the angle of which the person is looking at the object is 45 degrees.

so ni = 1.00029 (air)
nR = 1.33 (water)

so is this just a matter of (2m x 1.33) + 1m = 3.66m actual depth? :confused:

~Amy
 
ok.. did some reading and looks like i was wrong (surprise, surprise).

so i go:
sin 0i = (ni/nR) x sin 45 degrees
sin 0i = (1.00029/1.33) x sin 45 degrees
sin 0i = 0.5318
0i = 32.1 degrees

Dactual = 2.0/tan 32.1 degrees
= 3.18 meters

opinions?

~Amy
 
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