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

• bobbo7410
In summary, the conversation discusses the use of a laser with a wavelength of 544 x 10^-7 m in air for eye surgery. The index of refraction of the fluid in the eye is 1.43 and the goal is to calculate the frequency of the laser waves in the eye. The solution involves finding the velocity of the laser in air and in the eye, which are different due to the change in medium. However, the frequency remains constant in both media. This is because frequency is a measure of the number of waves passing a certain point per second, and this number cannot change regardless of the medium. Therefore, the frequency of the laser waves in the eye is the same as that in air.
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

Last edited:
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.

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..

Last edited:
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??

## 1. What is the index of refraction?

The index of refraction is a measure of how much a material can bend or slow down light as it passes through it. It is represented by the symbol "n" and is specific to each material.

## 2. How do you calculate the index of refraction?

The index of refraction can be calculated by dividing the speed of light in a vacuum by the speed of light in the material. This can be represented by the equation n = c/v, where c is the speed of light in a vacuum and v is the speed of light in the material.

## 3. What is the frequency of light?

The frequency of light is the number of waves that pass a certain point per second. It is usually measured in Hertz (Hz) or cycles per second.

## 4. How is frequency related to the index of refraction?

The frequency of light does not change as it passes through a material with a different index of refraction. However, the speed and wavelength of light will change, which can affect how the light is perceived.

## 5. How do I solve for frequency when there are two unknown variables?

In order to solve for frequency when there are two unknown variables, you will need to have at least two equations that relate the variables together. You can then use algebraic methods, such as substitution or elimination, to solve for the unknown variables and find the frequency. It may also be helpful to rearrange equations to solve for the desired variable before plugging in values.

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