Question

The speed of electromagnetic waves (which include visible light, radio and X-rays) in vacuum is $3 \times {10^8}\dfrac{m}{{\sec }}$ . Wavelengths of visible light waves range from about $400\,nm$ in the violet to about $700\,nm$ in the red. What is the range of frequencies of these waves?

Answer 1

Hint:A varying electric field will produce a varying magnetic field and vice versa. The two waves are perpendicular to each other and they propagate in a direction which is perpendicular to both the waves. They are mutually perpendicular. This is known as an electromagnetic wave. The wave has a few characteristics like frequency, speed, wavelength. They all are related by different formulae. We know that frequency is inversely proportional to wavelength. Using the above relation, we will find the minimum and maximum frequencies of the EM wave.

Complete step by step answer:
We know that the speed of electromagnetic waves in vacuum is $3 \times {10^8}\dfrac{m}{{\sec }}$. The formula that relates the wavelength and frequency is:
$f = \dfrac{v}{\lambda }$
Where $f$ = frequency, $v$ is the velocity of the wave, which in vacuum is equal to speed of the light and $\lambda $ is the wavelength of the wave.

When frequency is maximum, the wavelength is minimum. When frequency is minimum, the wavelength is maximum. The wavelength range given is: $(400\,nm,700\,nm)$.Therefore, ${\lambda _{\min }} = \dfrac{c}{{{f_{\max }}}}$ and ${\lambda _{\max }} = \dfrac{c}{{{f_{\min }}}}$
When we use the formula, we need to make sure that all the quantities are SI or CGS.
$1\,nm = 1 \times {10^{ - 9}}m$

Now substituting them in the formula, we get:
${f_{\min }} = \dfrac{{3 \times {{10}^8}}}{{700 \times {{10}^{ - 9}}}}$
$\Rightarrow {f_{\min }} = \dfrac{{3 \times {{10}^8}}}{{700 \times {{10}^{ - 9}}}} \\
\Rightarrow {f_{\min }}= 0.00428 \times {10^{17}} = 4.28 \times {10^{14}}Hz$
$\Rightarrow {f_{\max }} = \dfrac{{3 \times {{10}^8}}}{{400 \times {{10}^{ - 9}}}}$
$\Rightarrow f_{\max } = \dfrac{{3 \times {{10}^8}}}{{400 \times {{10}^{ - 9}}}} \\
\Rightarrow f_{\max } = 0.0075 \times 10^{17}\\
\therefore f_{\max } = 7.5 \times 10^{14}Hz$

Hence, the range of frequencies is \[(4.28 \times {10^{14}}Hz,7.5 \times {10^{14}}Hz)\].

Note:We need to remember the formulae relating the wavelength and frequency.The frequency is inversely proportional to wavelength. So, when one is maximum, the other one is minimum. One should avoid calculation mistakes while solving. A simple mistake can give a wrong answer. We also need to remember the nanometer and meter conversions.One should always express all the quantities in the same units, either SI or CGS, preferably SI.