Answer:
The answer is "12388.17"
Explanation:
[tex]l = 2.1 cm = 2.1 \times 10^{-2}\ m\\\\k = 1.00059\\\\\eta = \sqrt{k}= \sqrt{1.00059}\\\\\lambda = 663 nm = 663 \times 10^{-9}\ m[/tex]
Users now know that perhaps the number of fingers the shift is provided when a path difference [tex]\Delta d[/tex] are inserts between both the two arms
[tex]N = \frac{\Delta d}{\lambda}[/tex]
The optical pull-up in the arm is initially given by
[tex]d = 2\eta l[/tex]
Its new length of the different sense as the reflection coefficient adjustments between [tex]\eta[/tex] (air) and 1 so if we evacuate air from of the arm (vacuum).
The new length of a path is therefore
[tex]d'' = 2l[/tex]
Therefore, the different path
[tex]\Delta d=d-d''=2l(\eta -1)[/tex]
So, The fringe shifts number are
[tex]N= \frac{2l(\eta -1)}{\lambda}[/tex]
[tex]= \frac{2 \times 2.1 \times 10^{-2} (\sqrt{1.00059}-1)}{663 \times 10^{-9}}\\\\=12388.17[/tex]
