Interferometer, Index of refraction of a gas

[s_stylie0728]

Homework Statement

One of the beams of an interferometer, as seen in the figure below, passes through a small glass container containing a cavity D = 1.34 cm deep. When a gas is allowed to slowly fill the container, a total of 222 dark fringes are counted to move past a reference line. The light used has a wavelength of 598 nm. Calculate the index of refraction of the gas, assuming that the interferometer is in vacuum

*Similar to Michelson-Morley Experiment*

Homework Equations

Based off of assumption:
Snells Law: n1*sin(theta1)=N2*SIN(THETA2)
T=1/f=[lamb]/c
t1=(2*L1/c)*(1/sqrt(1-(v^2/c^2)))

The Attempt at a Solution

One variable I tried to solve for was t1 by finding the frequency, but that was unsuccessful and to be honest I'm not even sure on how to attempt this problem. If anyone has a hint as to where I could start, I would really appreciate it!

 

[Jhero]

Thank you for your question. I would like to help you in solving this problem. First, let's review the information given to us:

1. The interferometer has a small glass container with a cavity of depth D = 1.34 cm.
2. When a gas is slowly filled in the container, a total of 222 dark fringes are counted to move past a reference line.
3. The light used has a wavelength of 598 nm.
4. The interferometer is assumed to be in vacuum.

Based on this information, we can use the equation for the index of refraction, n, which is defined as the ratio of the speed of light in vacuum to the speed of light in the medium.

n = c/v

Where:
c = speed of light in vacuum (3 x 10^8 m/s)
v = speed of light in the medium

To find the speed of light in the medium, we can use the formula for the time delay, t1, which is the time it takes for the light to travel through the interferometer.

t1 = (2L/c)*(1/sqrt(1-(v^2/c^2)))

Where:
L = distance traveled by light in the interferometer
c = speed of light in vacuum
v = speed of light in the medium

We can rearrange the equation to solve for v:

v = c/sqrt(1-(c*t1/2L)^2)

Now, we need to find the value of t1. We can use the formula for frequency, f, which is defined as the number of cycles per unit time.

f = 1/T

Where:
T = time period
f = frequency

We can rearrange the equation to solve for T:

T = 1/f

Since we are given the wavelength of the light, we can use the formula for time period, T, which is defined as the time it takes for one wavelength to pass a point.

T = [lambda]/c

Where:
[lambda] = wavelength
c = speed of light in vacuum

We can now substitute this into the equation for t1:

t1 = (2L/c)*(1/sqrt(1-([lambda]*f)^2))

We are given the value of [lambda] as 598 nm, which is equivalent to 5.98 x 10^-7 m. We also know that