Answer to Question #155631 in Optics for kaashfi

Question #155631
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=?
1
Expert's answer
2021-01-14T10:37:24-0500

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"



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