Answer:
v = 3470 m/s
Step-by-step explanation:
given data:
suppose n be the number of fringes shifted which is equal to 0.005
wavelength of light is given as 589 nm
difference light path is given as [tex]\lambda n = \Delta d[/tex]
but we know that
[tex]\Delta d = c(\Delta t' -\Delta t)[/tex]
[tex](\Delta t' -\Delta t) =\frac{v^2(l1+l2)}{c^3}[/tex]
but in the question we have given that
l1=l2 = 11 m
therefore we have
[tex]\lambda n = c\frac{v^2(2l)}{c^3}[/tex]
solving for v, we get
[tex]v = c\sqrt {\frac{\lambda n}{2l}}[/tex]
[tex]v = 3*10^8\sqrt {\frac{589*10^{-9}*0.005}{2*11}}[/tex]
v = 3470 m/s
