Yellow light ( wavelength = 589.00 nm) illuminates a michelson interferometer. How many bright fringes will be counted as the mirror is moved through 1.00 cm?
Answers
Explanation:
Solution For a 630-nm red laser light, and for each fringe crossing \left(m=1\right), the distance traveled by {\text{M}}_{2} if you keep {\text{M}}_{1} fixed is
\text{Δ}d=m\frac{{\lambda }_{0}}{2}=1\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{630\phantom{\rule{0.2em}{0ex}}\text{nm}}{2}=315\phantom{\rule{0.2em}{0ex}}\text{nm}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}0.315\phantom{\rule{0.2em}{0ex}}\mu \text{m}.
SignificanceAn important application of this measurement is the definition of the standard meter. As mentioned in Units and Measurement, the length of the standard meter was once defined as the mirror displacement in a Michelson interferometer corresponding to 1,650,763.73 wavelengths of the particular fringe of krypton-86 in a gas discharge tube.
Measuring the Refractive Index of a Gas In one arm of a Michelson interferometer, a glass chamber is placed with attachments for evacuating the inside and putting gases in it. The space inside the container is 2 cm wide. Initially, the container is empty. As gas is slowly let into the chamber, you observe that dark fringes move past a reference line in the field of observation. By the time the chamber is filled to the desired pressure, you have counted 122 fringes move past the reference line. The wavelength of the light used is 632.8 nm. What is the refractive index of this gas?
Pictures shows a schematic of a set-up utilized to measure the refractive index of a gas. The glass chamber with a gas is placed in the Michelson interferometer between the half-silvered mirror M and mirror M1. The space inside the container is 2 cm wide.
Strategy The m=122 fringes observed compose the difference between the number of wavelengths that fit within the empty chamber (vacuum) and the number of wavelengths that fit within the same chamber when it is gas-filled. The wavelength in the filled chamber is shorter by a factor of n, the index of refraction.
Solution The ray travels a distance t=2\phantom{\rule{0.2em}{0ex}}\text{cm} to the right through the glass chamber and another distance t to the left upon reflection. The total travel is L=2t. When empty, the number of wavelengths that fit in this chamber is
{N}_{0}=\frac{L}{{\lambda }_{0}}=\frac{2t}{{\lambda }_{0}}
where {\lambda }_{0}=632.8\phantom{\rule{0.2em}{0ex}}\text{nm} is the wavelength in vacuum of the light used. In any other medium, the wavelength is \lambda ={\lambda }_{0}\text{/}n and the number of wavelengths that fit in the gas-filled chamber is
N=\frac{L}{\lambda }=\frac{2t}{{\lambda }_{0}\text{/}n}.
The number of fringes observed in the transition is
\begin{array}{cc}m& =N-{N}_{0},\hfill \\ & =\frac{2t}{{\lambda }_{0}\text{/}n}-\frac{2t}{{\lambda }_{0}},\hfill \\ & =\frac{2t}{{\lambda }_{0}}\left(n-1\right).\hfill \end{array}
Solving for \left(n-1\right) gives
n-1=m\left(\frac{{\lambda }_{0}}{2t}\right)=122\left(\frac{632.8\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-9}\phantom{\rule{0.2em}{0ex}}\text{m}}{2\left(2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-2}\phantom{\rule{0.2em}{0ex}}\text{m}\right)}\right)=0.0019
and n=1.0019.
Significance The indices of refraction for gases are so close to that of vacuum, that we normally consider them equal to 1. The difference between 1 and 1.0019 is so small that measuring it requires a correspondingly sensitive technique such as interferometry. We cannot, for example, hope to measure this value using techniques based simply on Snell’s law.