Question

The wavelength of visible light ranges from 400 nm to 700 nm.

A) Find the ranges of this light's frequency.
Answer in the order indicated. Separate answers using comma. (answer in Hz)

B) Find the ranges of this light's angular frequency.
Answer in the order indicated. Separate answers using comma. (answer in rad/s)

C) Find the ranges of this light's wave number.
Answer in the order indicated. Separate answers using comma. (answer in rad/m)

Answer 1

Concepts and reason

The concept used to solve the problem is frequency of light, angular frequency and wave number.

Initially, the frequency of light is calculated from the wavelength of the visible light. Later, the angular frequency is calculated from the visible light. Finally, the wave number is calculated from the wavelength of light.

Fundamentals

The relation between the frequency of light and wavelength of light is given as:

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

Here, c is the velocity of light, $\lambda$ is the wavelength of light and f is the frequency of light.

The angular frequency of light is given as:

$\omega = 2\pi f$

Here, f is the frequency of light.

The wavelength of the light and wave number are related as:

$k = \frac{{2\pi }}{\lambda }$

Here, $\lambda$ is the wavelength and $k$ is the wavenumber.

The frequency of light is given as:

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

The frequency is inversely proportional to wavelength.

Substitute 700nm for $\lambda$ , $3 \times {10^8}\;{\rm{m/s}}$ for c in $f = \frac{c}{\lambda }$ and find the frequency.

$\begin{array}{c}\\f = \frac{c}{\lambda }\\\\ = \frac{{3 \times {{10}^8}\;{\rm{m/s}}}}{{700\;{\rm{nm}}}}\\\\ = \frac{{3 \times {{10}^8}\;{\rm{m/s}}}}{{700 \times {{10}^{ - 9}}{\rm{m}}}}\\\\ = 4.28\; \times {10^{14}}{\rm{Hz}}\\\end{array}$

The frequency is $4.28\; \times {10^{14}}{\rm{Hz}}$ .

Substitute 400nm for $\lambda$ , $3 \times {10^8}\;{\rm{m/s}}$ for c in $f = \frac{c}{\lambda }$ and find the frequency.

$\begin{array}{c}\\f = \frac{c}{\lambda }\\\\ = \frac{{3 \times {{10}^8}\;{\rm{m/s}}}}{{400\;{\rm{nm}}}}\\\\ = \frac{{3 \times {{10}^8}\;{\rm{m/s}}}}{{400 \times {{10}^{ - 9}}{\rm{m}}}}\\\\ = 7.5\; \times {10^{14}}{\rm{Hz}}\\\end{array}$

The frequency is $7.5\; \times {10^{14}}{\rm{Hz}}$ .

The frequency is between $4.28\; \times {10^{14}}{\rm{Hz}}$ and $7.5\; \times {10^{14}}{\rm{Hz}}$ .

The maximum frequency is $7.5\; \times {10^{14}}{\rm{Hz}}$ and the minimum frequency is $4.28\; \times {10^{14}}{\rm{Hz}}$ .

The angular frequency of light is given as:

$\omega = 2\pi f$

Here, f is the frequency of light.

The angular frequency of the light is proportional to frequency of light.

The maximum angular frequency is given as:

${\omega _{\max }} = 2\pi {f_{\max }}$

Substitute $7.5\; \times {10^{14}}{\rm{Hz}}$ for ${f_{\max }}$ and find ${\omega _{\max }}$ .

$\begin{array}{c}\\{\omega _{\max }} = 2\pi {f_{\max }}\\\\ = 2\pi \left( {7.5\; \times {{10}^{14}}{\rm{Hz}}} \right)\\\\ = 4.71 \times {10^{15}}\;{\rm{rad/s}}\\\end{array}$

The minimum angular frequency is given as:

${\omega _{\min }} = 2\pi {f_{\min }}$

Substitute $4.28\; \times {10^{14}}{\rm{Hz}}$ for ${f_{\min }}$ and find ${\omega _{\min }}$ .

$\begin{array}{c}\\{\omega _{\min }} = 2\pi {f_{\min }}\\\\ = 2\pi \left( {4.28\; \times {{10}^{14}}{\rm{Hz}}} \right)\\\\ = 2.68 \times {10^{15}}\;{\rm{rad/s}}\\\end{array}$

The maximum angular frequency is $4.71 \times {10^{15}}\;{\rm{rad/s}}$ and the minimum angular frequency is $2.68 \times {10^{15}}\;{\rm{rad/s}}$ .

The wavelength of the light and wave number are related as:

$k = \frac{{2\pi }}{\lambda }$

Here, $\lambda$ is the wavelength and $k$ is the wavenumber.

The wavenumber of light is inversely proportional to wavelength. The wavenumber is calculated as follows:

$k = \frac{{2\pi }}{\lambda }$

Substitute 700nm for $\lambda$ and find the wave number.

$\begin{array}{c}\\k = \frac{{2\pi }}{\lambda }\\\\ = \frac{{{\rm{2\pi }}}}{{{\rm{700 \times 1}}{{\rm{0}}^{{\rm{ - 9}}}}{\rm{m}}}}\\\\ = 8.97 \times {10^6}\;{\rm{rad/m}}\\\end{array}$

Substitute 400nm for $\lambda$ and find the wave number.

$\begin{array}{c}\\k = \frac{{2\pi }}{\lambda }\\\\ = \frac{{{\rm{2\pi }}}}{{{\rm{400 \times 1}}{{\rm{0}}^{{\rm{ - 9}}}}{\rm{m}}}}\\\\ = 1.57 \times {10^7}\;{\rm{rad/m}}\\\end{array}$

The wave numbers range from $8.97 \times {10^6}\;{\rm{rad/m}}$ to $1.57 \times {10^7}\;{\rm{rad/m}}$ .

Ans:

The maximum frequency is $7.5\; \times {10^{14}}{\rm{Hz}}$ and the minimum frequency is $4.28\; \times {10^{14}}{\rm{Hz}}$

The maximum angular frequency is $4.71 \times {10^{15}}\;{\rm{rad/s}}$ and the minimum angular frequency is $2.68 \times {10^{15}}\;{\rm{rad/s}}$ .

The maximum wave number is $1.57 \times {10^7}\;{\rm{rad/m}}$ and the minimum wave number is

$8.97 \times {10^6}\;{\rm{rad/m}}$ .