Question

If the frequencies of an ultrasonic wave and an electromagnetic wave are same, then:
A. their wavelength will be same
B. wavelength of electromagnetic wave will be less than that of ultrasonic wave
C. wavelength of electromagnetic wave will be more than that of ultrasonic wave
D. the wavelengths of both will be nearly equal

Answer 1

The electromagnetic and ultrasonic waves' velocity is equal to the product of their respective wavelengths and frequencies. We will substitute the frequency of both the waves equal in velocities' expressions to determine the relationship between their wavelengths.

Complete step by step answer:
We have classified sound waves based on their frequency range. An ultrasonic wave is a sound wave having a frequency more than $20{\rm{ kHz}}$ which is higher than the audible frequency range of human beings.

Electromagnetic waves are the rays of light, and as the name suggests, they are obtained as the result of a combination of vibrations produced due to magnetic and electric fields.

The expression for speed of the given ultrasonic wave is:
${v_1} = {\lambda _1}{\nu _1}$……(1)
Here ${\lambda _1}$ is the wavelength and ${\nu _1}$ is the frequency of the ultrasonic wave.

The expression for the speed of an electromagnetic wave is:
${v_2} = {\lambda _2}{\nu _2}$……(2)
Here ${\lambda _2}$ is the wavelength and ${\nu _2}$ is the frequency of the electromagnetic wave.

Divide equation (1) and equation (2).
$\dfrac{{{v_1}}}{{{v_2}}} = \dfrac{{{\lambda _1}{\nu _1}}}{{{\lambda _2}{\nu _2}}}$……(4)

It is given that the frequency of the ultrasonic wave is equal to the frequency of the electromagnetic wave.
${\nu _1} = {\nu _2}$

Substitute ${\nu _2}$ for ${\nu _1}$ in equation (4).

\dfrac{{{v_1}}}{{{v_2}}} = \dfrac{{{\lambda _1}\left( {{\nu _2}} \right)}}{{{\lambda _2}{\nu _2}}}\\\ \dfrac{{{\lambda _1}}}{{{\lambda _2}}} = \dfrac{{{v_1}}}{{{v_2}}} \end{array}$$……(5) We know that the speed of the ultrasonic wave is less than the speed of the electromagnetic wave. $$\begin{array}{l} {v_1} < {v_2}\\\ \dfrac{{{v_1}}}{{{v_2}}} < 1 \end{array}$$ Substitute less than $$1$$ for $$\dfrac{{{v_1}}}{{{v_2}}}$$ in equation (5). $$\begin{array}{l} \dfrac{{{\lambda _1}}}{{{\lambda _2}}} < 1\\\ {\lambda _2} > {\lambda _1} \end{array}$$ Therefore, we can say that the electromagnetic wave wavelength is more than the wavelength of the ultrasonic wave, keeping their frequencies equal **So, the correct answer is “Option C”.** **Note:** The relation between wavelengths of electromagnetic wave and ultrasonic wave takes care of less than or greater than sign. Alternatively, we could have determined the ratio of both the waves' velocities and substituted that value in equation (4).