Answer to Question #155610 in Optics for Huzaifa Rehman

Question

If the mirror separation d = 50 cm changes by 2 nm, what would be the change in relative frequency? If the laser frequency is νL = 4 x 1014s-1, what would be the value of ∆νL=?

Answer 1

$\Delta f=\frac{c}{2L}=\frac{\left(3\times10^8\frac{m}{s}\right)}{\left(2\times2\times10^{-9}m\right)}=7.5\times10^{16}s^{-1}$

$vL=4\times10^{14}s^{-1}$

$\Delta vL=(7.5\times10^{16}s^{-1})-(4\times10^{14}s^{-1})=7.46x10^{16}$