Index of refraction and electromagnetic radiation

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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