Chapter 17, Problem 1CQ

The frequency of a light wave in air is 5.3 × 10¹⁴ Hz. Is the frequency of this wave higher, lower, or the same after the light enters a piece of glass?

To determine

To find: The frequency of light wave is higher, lower or the same after the light enters a piece of glass.

Answer to Problem 1CQ

The frequency of light wave is same after the light enters a piece of glass.

Explanation of Solution

Given data:

The frequency of light wave is $5.3\times {10}^{14}\text{Hz}$.

Formula used:

Write the expression for refractive index.

$n=\frac{c}{v}$ (1)

Here,

$n$ is index of refraction,

$c$ is the speed of light in vacuum, and

$v$ is the speed of light in a material.

Write the expression for speed.

$v=f\lambda$ (2)

Here,

$f$ is the frequency, and

$\lambda$ is wavelength.

Explanation:

Draw the light wave travels from air to glass as shown in Figure 1.

Refer to Table 17.1 in the textbook, the value of index of refraction for glass and air is $1.50$ and $1.0003$ respectively.

Modify equation (1) for air as follows.

$n_{\text{air}}=\frac{c}{v_{\text{air}}}$ (3)

Modify equation (2) for air as follows.

$v_{\text{air}}=f_{\text{air}}{\lambda }_{\text{air}}$

Substitute $f_{\text{air}}{\lambda }_{\text{air}}$ for $v_{\text{air}}$ in equation (3),

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

Rearrange the equation.

${\lambda }_{\text{air}}=\frac{c}{n_{\text{air}}{f}_{\text{air}}}$ (4)

Modify equation (1) for glass as follows.

$n_{\text{glass}}=\frac{c}{v_{\text{glass}}}$ (5)

Modify equation (2) for air as follows.

$v_{\text{glass}}=f_{\text{glass}}{\lambda }_{\text{glass}}$

Substitute $f_{\text{glass}}{\lambda }_{\text{glass}}$ for $v_{\text{glass}}$ in equation (5),

$n_{\text{glass}}=\frac{c}{f_{\text{glass}}{\lambda }_{\text{glass}}}$

Rearrange the equation as follows.

${\lambda }_{\text{glass}}=\frac{c}{n_{\text{glass}}{f}_{\text{glass}}}$ (6)

Here, the value of $c$ is $3\times {10}^8\frac{\text{m}}{\text{s}}$.

Substitute $3\times {10}^8\frac{\text{m}}{\text{s}}$ for $c$, $1.0003$ for $n_{\text{air}}$ and $5.3\times {10}^{14}\text{Hz}$ for $f_{\text{air}}$ in equation (4),

$\begin{array}{c}{\lambda }_{\text{air}}=\frac{3\times {10}^8\frac{\text{m}}{\text{s}}}{\left(1.0003\right)\left(5.3\times {10}^{14}\text{Hz}\right)}\\ =5.65\times {10}^{-7}\text{m}\\ \cong 5.7\times {10}^{-7}\text{m}\end{array}$

Substitute $3\times {10}^8\frac{\text{m}}{\text{s}}$ for $c$, $1.5$ for $n_{\text{glass}}$ and $5.3\times {10}^{14}\text{Hz}$ for $f_{\text{glass}}$ in equation (6),

$\begin{array}{c}{\lambda }_{\text{glass}}=\frac{3\times {10}^8\frac{\text{m}}{\text{s}}}{\left(1.5\right)\left(5.3\times {10}^{14}\text{Hz}\right)}\\ =3.7\times {10}^{-7}\text{m}\end{array}$

Find the value of $n_{\text{air}}{\lambda }_{\text{air}}$.

$\begin{array}{c}{n}_{\text{air}}{\lambda }_{\text{air}}=\left(1.0003\right)\left(5.7\times {10}^{-7}\text{m}\right)\\ =5.701\times {10}^{-7}\text{m}\\ \cong 5.7\times {10}^{-7}\text{m}\end{array}$

Find the value of $n_{\text{glass}}{\lambda }_{\text{glass}}$.

$\begin{array}{c}{n}_{\text{glass}}{\lambda }_{\text{glass}}=\left(1.5\right)\left(3.7\times {10}^{-7}\text{m}\right)\\ =5.55\times {10}^{-7}\text{m}\\ \cong 5.6\times {10}^{-7}\text{m}\end{array}$

Write the relation between $n_{\text{air}}{\lambda }_{\text{air}}$ and $n_{\text{glass}}{\lambda }_{\text{glass}}$ using obtained values.

$n_{\text{air}}{\lambda }_{\text{air}}\cong n_{\text{glass}}{\lambda }_{\text{glass}}$

Re-arrange equation (4) as follows.

$f_{\text{air}}=\frac{c}{n_{\text{air}}{\lambda }_{\text{air}}}$ (7)

Re-arrange equation (6) as follows.

$f_{\text{glass}}=\frac{c}{n_{\text{glass}}{\lambda }_{\text{glass}}}$ (8)

Divide equation (7) by (8).

$\begin{array}{c}\frac{f_{\text{air}}}{f_{\text{glass}}}=\frac{\left(\frac{c}{n_{\text{air}}{\lambda }_{\text{air}}}\right)}{\left(\frac{c}{n_{\text{glass}}{\lambda }_{\text{glass}}}\right)}\\ =\frac{n_{\text{glass}}{\lambda }_{\text{glass}}}{n_{\text{air}}{\lambda }_{\text{air}}}\end{array}$

Substitute $n_{\text{glass}}{\lambda }_{\text{glass}}$ for $n_{\text{air}}{\lambda }_{\text{air}}$,

$\begin{array}{l}\frac{f_{\text{air}}}{f_{\text{glass}}}=\frac{n_{\text{glass}}{\lambda }_{\text{glass}}}{n_{\text{glass}}{\lambda }_{\text{glass}}}\\ \frac{f_{\text{air}}}{f_{\text{glass}}}=1\\ f_{\text{air}}=f_{\text{glass}}\end{array}$

Conclusion:

Hence, the frequency of light wave is same after the light enters a piece of glass.