Question

The wavelength of sodium light in air is 589 nm. (a) Find its frequency in air. (b) Find its wavelength in water (refractive index = 1.33). (c) Find its frequency in water. (d) Find its speed in water.

Answer 1

Given:
Wavelength of sodium light in air, ${\lambda }_{\mathrm{a}}=589\mathrm{nm}=589\times {10}^{-9}\mathrm{m}$
Refractive index of water, μ_w= 1⋅33
We know that $f=\frac{c}{\lambda }$,
$\mathrm{where}c=\mathrm{speed}\mathrm{of}\mathrm{light}=3\times {10}^8\mathrm{m}/\mathrm{s}$
f = frequency
λ = wavelength
(a) Frequency in air, $f_{\mathrm{air}}=\frac{c}{{\lambda }_{\mathrm{a}}}$
$$\begin{gathered} f_{\mathrm{air}}=\frac{3\times {10}^8}{589\times {10}^{-9}} \\ =5.09\times {10}^{14}\mathrm{Hz} \end{gathered}$$

(b)
Let wavelength of sodium light in water be ${\lambda }_{\mathrm{w}}$.
We know that
$\frac{{\mathrm{\mu }}_{\mathrm{a}}}{{\mathrm{\mu }}_{\mathrm{w}}}=\frac{{\lambda }_{\mathrm{\omega }}}{{\lambda }_{\mathrm{a}}}$,
where μ_a is the refractive index of air which is equal to 1 and
λ_w is the wavelength of sodium light in water.
$$\begin{gathered} \Rightarrow \frac{1}{1.33}=\frac{{\lambda }_{\mathrm{\omega }}}{589\times {10}^{-9}} \\ \Rightarrow {\lambda }_{\mathrm{\omega }}=443\mathrm{nm} \end{gathered}$$

(c) Frequency of light does not change when light travels from one medium to another.
$$\begin{gathered} \therefore f_{\mathrm{\omega }}=f_{\mathrm{a}} \\ =5.09\times {10}^{14}\mathrm{Hz} \end{gathered}$$

(d) Let the speed of sodium light in water be ${\nu }_{\mathrm{\omega }}$
and speed in air, v_a = c.
Using $\frac{{\mu }_{\mathrm{a}}}{{\mu }_{\mathrm{\omega }}}=\frac{{\nu }_{\mathrm{\omega }}}{{\nu }_{\mathrm{a}}}$, we get:
$$\begin{gathered} {\nu }_{\mathrm{\omega }}=\frac{{\mu }_{\mathrm{a}}c}{{\mu }_{\mathrm{\omega }}} \\ =\frac{3\times {10}^8}{\left(1.33\right)}=2.25\times {10}^8\mathrm{m}/\mathrm{s} \end{gathered}$$