Chapter 43: Problem 39

Question: Light of wavelength $\lambda~=~612~{\rm nm}$ in a vacuum travels $d~=~1.57~{\rm\mu m}$ in a medium of refractive index n = 1.51. Find the wavelength in the medium, the optical path length and the phase difference after moving that distance wrt light traveling the same distance in a vacuum.

Solution: The frequency in the medium is the same as in the vacuum, and so the wavelength must be reduced by the ratio of speeds, hence $\lambda_2~=~\lambda/n~=~405~{\rm nm}$. The optical path is the distance the light would have traveled in the time it took to travel the distance in the medium. Therefore, it is simply the distance traveled in the medium, multipled by the ratio of refractive indices, $L~=~d n~=~ 2.37~{\rm\mu m}$. The phase difference upon after traveling dis

$$\phi~=~(k_1-k_2) d~=~ 2\pi d \left( {1\over\lambda_1}-{1\over\lambda_2}\right)$$

$\phi~=~8.23~{\rm rad}~=~471.6^o$, and therefore a phase difference of 111.6^o.