Question

Determine the change in wavelength of light during its passage from air to glass. If the refractive index of glass with respect to air is $1\cdot 5$ and the frequency of light is $4\times {10}^{14}Hz$. Find the wave number of light In glass. [Velocity of light air $(c=3\times {10}^{8}m/s)$]

Answer 1

Given : ${\mu }_8=1.5;n=4\times {10}^{14}Hz;c=3\times {10}^{8}m/s$

The wavelength of light incident on glass from air is

$x_a=\frac{c}{n}=\frac{3\times {10}^8}{4\times {10}^{14}}=7.5\times {10}^{-7}m=7500\times {10}^{-10}=7500A$

Now, the velocity of light in glass is given from its refractive index as

${\mu }_g=\frac{c}{v_g}$

We also know that velocity is product of frequency and wavelength.

$\therefore {\mu }_g=\frac{c}{v_g}=\frac{n{\lambda }_{\sigma }}{n{\lambda }_g}=\frac{{\lambda }_{\sigma }}{{\lambda }_g}$

$\therefore {\lambda }_g=\frac{{\lambda }_{\sigma }}{{\lambda }_g}=\frac{7500}{1.5}=5000\stackrel{o}{A}$

Therefore, the difference in wavelength is

${\lambda }_{\sigma }-{\lambda }_g=7500-5000=2500\stackrel{o}{A}$

The wave number is the reciprocal of the wavelength.

Therefore, the wave number in glass is

${\overline{\lambda }}_g=\frac{1}{{\lambda }_g}$

$\therefore {\overline{\lambda }}_g=\frac{1}{5\times {10}^{-7}}=2\times {10}^6{m}^{-1}$