Respuesta :
Explanation:
When Michelson-Morley apparatus is turned through [tex]90^{o}[/tex] then position of two mirrors will be changed. The resultant path difference will be as follows.
[tex]\frac{lv^{2}}{\lambda c^{2}} - (-\frac{lv^{2}}{\lambda c^{2}}) = \frac{2lv^{2}}{\lambda c^{2}}[/tex]
Formula for change in fringe shift is as follows.
n = [tex]\frac{2lv^{2}}{\lambda c^{2}}[/tex]
[tex]v^{2} = \frac{n \lambda c^{2}}{2l}[/tex]
v = [tex]\sqrt{\frac{n \lambda c^{2}}{2l}}[/tex]
According to the given data change in fringe is n = 1. The data is Michelson and Morley experiment is as follows.
l = 11 m
[tex]\lambda = 5.9 \times 10^{-7} m[/tex]
c = [tex]3.0 \times 10^{8}[/tex] m/s
Hence, putting the given values into the above formula as follows.
v = [tex]\sqrt{\frac{n \lambda c^{2}}{2l}}[/tex]
= [tex]\sqrt{\frac{1 \times (5.9 \times 10^{-7} m) \times (3.0 \times 10^{8})^{2}}{2 \times 11 m}}[/tex]
= [tex]2.41363 \times 10^{9} m/s[/tex]
Thus, we can conclude that velocity deduced is [tex]2.41363 \times 10^{9} m/s[/tex].
