Index of refraction, calculate frequency. ~ stuck at 2 unknown variables

[bobbo7410]

Homework Statement

An laser is used of wavelength 544 x 10^-7 m (in air) for eye surgery. The index of refraction of the fluid in an eye is n = 1.43. Calculate the frequency of the laser waves in the eye.

n = 1.43
L1=544 x 10^-7
L2=
f =

Homework Equations

n=L1/L2

L1 being the wavelength in "air"
L2 being the wavelength in the eye

f=v/L

L=v/f

The Attempt at a Solution

so, given n and L1 its simple to find L2, so I now know the wavelength in the eye. Yet I can't seem to derive the frequency.

so:

n=L1/L2

1.43= (5.44 x 10^-7) / (v/f)

the frequency remains constant for both wavelengths.

so simply I would have to find the velocity of L1 or L2 that would be the frequency "within the eye"

v1/L1=v2/L2

the velocity is not the same so I now have 2 unknown variables.

I KNOW this isn't hard, my minds just shot right now

 

[Dick]

You already said the correct answer. The frequency is the same. The wavelengths and velocities are different, but they both change by the same factor.

 

[bobbo7410]

Thanks Dick,

I had though that, but then that same factor that they change by is simply equal to n. Using n for the frequency wouldn't create the correct answer. I'm just not thinking right.

If you or anyone else could elaborate it would be greatly appreciated.

I know this isn't complicated and I'm embarrassed to continue asking when I'm sure this is so simple..

 

[Dick]

Air has basically n=1. f=v/lambda. In a different medium v->v/n, lambda->lambda/n. The frequency is unchanged. This is also sort of obvious. A thing that measures frequency is counting waves as they pass. If it's located outside of the eye it counts f waves per second. If it's inside the eye it must also count f waves per second. Otherwise, some 'waves' are getting lost somewhere. Where could they go??