Chapter 37, Problem 14SP

To determine

The frequency of light in vacuum and water and its speed and wavelength in water if the wavelength is $546\text{ nm}$ in airand the refractive index of water is $1.33$.

Answer to Problem 14SP

Solution:

$549\text{ THz}$, $549\text{ THz}$, $2.25\times {10}^8\text{ m/s}$, and $410\text{ nm}$.

Explanation of Solution

Given data:

The wavelength of light in air is $546\text{ nm}$.

The refractive index of water is $1.33$.

Formula used:

The formula for frequency of light in a medium is as follows:

$f=\frac{v}{\lambda }$

Here, $f$ is the frequency of light in the material, $v$ is the velocity of light in the medium, and $\lambda$ is the wavelength of light in the medium.

The formula for refractive index of a medium is

$n=\frac{c}{v}$

Here, $n$ is the refractive index of the medium, $c$ is the speed of light in vacuum and $v$ is the velocity of light in the medium .

Explanation:

Recall the expression for frequency of light in air as a medium:

$f_{air}=\frac{c}{{\lambda }_{air}}$

Here, $f_{air}$ is the frequency of light in air, $c$ is the speed of light in a vacuum, and ${\lambda }_{air}$ is the wavelength of light in the air as a medium .

Substitute $3\times {10}^8\text{ m/s}$ for $c$ and $\text{546 nm}$ for ${\lambda }_{air}$

$\begin{array}{c}{f}_{air}=\frac{3\times {10}^8\text{ m/s}}{546\text{ nm}}\\ =\frac{3\times {10}^8{\text{ ms}}^{-1}}{546\text{ nm}\left(\frac{{10}^{-9}\text{ m}}{1\text{ nm}}\right)}\\ =5.49\times {10}^{14}\text{ Hz}\left(\frac{{10}^{-12}\text{ THz}}{1\text{ Hz}}\right)\\ =549\text{ THz}\end{array}$

The frequency of light is the same in any medium. However, its wavelength and velocity are different in different mediums.

Hence, the frequency of light in water is $549\text{ THz}$ .

Recall the expression for refractive index of medium:

$n=\frac{c}{v}$

The above expression can be re-written, if the medium is water, as

$v_{\text{water}}=\frac{c}{n_{\text{water}}}$

Here, $n_{water}$ is the refractive index of water, $c$ is the speed of light in a vacuum, and $v_{water}$ is the velocity of light in water medium.

Substitute $3\times {10}^8\text{ m/s}$ for $c$ and $1.33$ for $v_{\text{water}}$

$\begin{array}{c}{v}_{\text{water}}=\frac{3\times {10}^8\text{ m/s}}{1.33}\\ =2.25\times {10}^8\text{ m/s}\end{array}$

Recall the expression for frequency of light in water as a medium:

$f_{\text{water}}=\frac{v_{\text{water}}}{{\lambda }_{\text{water}}}$

It is written for the wavelength of light in water as

${\lambda }_{\text{water}}=\frac{v_{\text{water}}}{f_{\text{water}}}$

Here, ${\lambda }_{\text{water}}$ is the wavelength in water, $v_{\text{water}}$ is the speed of light in water, and $f_{\text{water}}$ is the frequency of light in water.

Substitute $2.25\times {10}^8\text{ m/s}$ for $v_{\text{water}}$ and $5.49\times {10}^{14}\text{ Hz}$ for $f_{\text{water}}$

$\begin{array}{c}{\lambda }_{\text{water}}=\frac{2.25\times {10}^8\text{ m/s}}{5.49\times {10}^{14}\text{ Hz}}\\ =\frac{2.25\times {10}^8{\text{ ms}}^{-1}}{5.49\times {10}^{14}\text{ Hz}}\\ \approx 4.100\times {10}^{-7}\text{ m}\left(\frac{1\text{ nm}}{{10}^{-9}\text{ m}}\right)\\ \approx 410\text{ nm}\end{array}$

Conclusion:

The frequency of light in air is $549\text{ THz}$, the frequency of light in water is $549\text{ THz}$, the speed of light in water is $2.25\times {10}^8\text{ m/s}$, and the wavelength of light in water is $410\text{ nm}$ .