Answer to Question #155631 in Optics for kaashfi

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

Answer

Relative frequency is given

$\lambda=0.5m$

$\Delta \lambda=2nm$

Frequency

$\nu=\frac{c}{\lambda}=6\times10^8Hz$

Now relative frequency is given

$\frac{\Delta \nu}{\nu}=\frac{c\Delta \lambda}{\lambda^2}=\frac{3\times10^8\times2\times10^{-9}}{0.5\times0.5}$

$=12$

$\Delta \nu_L=12\times4 \times 10^{14}\\=48 \times 10^{14}Hz$