Question

problem 24.59
one of the beams of an interferometer (figure 1) passes through a small glass container containing a cavity 1.155 cm deep. when a gas is allowed to slowly fill the container, a total of 148 dark fringes are counted to move past a reference line. the light used has a wavelength of 543.1 nm.
figure
1 of 1
to mirror m₁
source
mₛ
glass container
1.155 cm
m₂
part a
calculate the index of refraction of the gas, assuming that the interferometer is in vacuum.
enter your answer using five decimal places.
n_gas =
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Explanation:

Step1: Recall the formula for fringe shift in interferometer

The number of fringe shifts \( N \) is related to the path difference. When a gas fills the cavity, the optical path length changes. The formula is \( N=\frac{2d(n - 1)}{\lambda} \), where \( d \) is the length of the cavity, \( n \) is the refractive index of the gas, and \( \lambda \) is the wavelength of light. We need to solve for \( n \).

Step2: Rearrange the formula to solve for \( n \)

From \( N=\frac{2d(n - 1)}{\lambda} \), we can rearrange it as \( n=1+\frac{N\lambda}{2d} \)

Step3: Convert units

First, convert \( d = 1.155\space cm=1.155\times 10^{- 2}\space m \) and \( \lambda=543.1\space nm = 543.1\times 10^{-9}\space m \), \( N = 148 \)

Step4: Substitute the values into the formula

Substitute \( N = 148 \), \( \lambda=543.1\times 10^{-9}\space m \), \( d = 1.155\times 10^{-2}\space m \) into \( n=1+\frac{N\lambda}{2d} \)

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Rounding to five decimal places, \( n\approx1.00348 \) (Wait, let's recalculate the division:

\( 148\times543.1=148\times500 + 148\times43.1=74000+6378.8 = 80378.8 \)

\( 2\times1.155\times 10^{-2}=2.31\times 10^{-2} \)

\( \frac{80378.8\times 10^{-9}}{2.31\times 10^{-2}}=\frac{80378.8}{2.31}\times 10^{-7}\approx34796\times 10^{-7}=3.4796\times 10^{-3} \)

Then \( n = 1+0.0034796=1.0034796\approx1.00348 \) (Wait, when we do \( 80378.8\div(2.31\times 10^{7}) \):

\( 2.31\times 10^{7}=23100000 \)

\( 80378.8\div23100000\approx0.0034796 \)

So \( n=1 + 0.0034796=1.0034796\approx1.00348 \) (But let's check the calculation again:

\( \frac{N\lambda}{2d}=\frac{148\times543.1\times 10^{-9}}{2\times1.155\times 10^{-2}}=\frac{148\times543.1}{2\times1.155}\times 10^{-7} \)

\( 2\times1.155 = 2.31 \)

\( 148\times543.1=80378.8 \)

\( 80378.8\div2.31\approx34796 \)

\( 34796\times 10^{-7}=0.0034796 \)

So \( n = 1+0.0034796 = 1.0034796\approx1.00348 \) (Wait, maybe I made a mistake in the first calculation. Let's use more precise steps:

\( \frac{148\times543.1\times 10^{-9}}{2\times1.155\times 10^{-2}}=\frac{148\times543.1}{2\times1.155}\times 10^{-7} \)

\( 2\times1.155 = 2.31 \)

\( 148\div2.31\approx64.069264 \)

\( 64.069264\times543.1\times 10^{-7} \)

\( 64.069264\times543.1 = 64\times543.1+0.069264\times543.1=34758.4+37.617 = 34796.017 \)

\( 34796.017\times 10^{-7}=0.0034796017 \)

So \( n=1 + 0.0034796017=1.0034796017\approx1.00348 \) (when rounded to five decimal places, 1.00348? Wait, 1.0034796017, the fifth decimal place is 9, the sixth is 6, so we round up the fifth decimal place: 1.00348? Wait, 1.0034796017:

The number is 1.00347 96017

Wait, the first decimal place: 0, second:0, third:3, fourth:4, fifth:7, sixth:9. So when rounding to five decimal places, we look at the sixth decimal place (9), which is greater than 5, so we add 1 to the fifth decimal place. So 7 + 1=8. So \( n = 1.00348 \)

Answer:

\( 1.00348 \) (Wait, let's check with more precise calculation:

\( N = 148 \), \( \lambda=543.1\space nm = 543.1e - 9\space m \), \( d=1.155\space cm = 0.01155\space m \)

\( n=1+\frac{148\times543.1e - 9}{2\times0.01155} \)

Calculate numerator of the fraction: \( 148\times543.1e - 9=148\times543.1\times10^{-9}=80378.8\times10^{-9} = 8.03788\times 10^{-5}\)

Denominator: \( 2\times0.01155 = 0.0231\)

Then \( \frac{8.03788\times 10^{-5}}{0.0231}\approx3.4796\times 10^{-3}\)

Then \( n = 1+0.0034796 = 1.0034796\), which is \( 1.00348 \) when rounded to five decimal places. )