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Question: Calculate the wavelength and frequency of a light wave whose period is \[2\times {{10}^{-10}}\sec \]...

Calculate the wavelength and frequency of a light wave whose period is 2×1010sec2\times {{10}^{-10}}\sec .

Explanation

Solution

Lightwave is a form of electromagnetic wave and electromagnetic waves are transverse waves, that is, their characteristic features are the crests and the troughs in their waveform. The distance between two consecutive crests or two consecutive troughs is known as the wavelength, whereas frequency is the number of waves passing a point in a given time interval, usually one second.

Complete step-by-step solution:
Data provided in the question,
The period of the wave (T)=2×1010sec(T)=2\times {{10}^{-10}}\sec
The velocity of any wave is related to the wavelength and frequency as
V=λfequation(1)V=\lambda f-----equation(1)
But in the above question, the specified wave is light, so we can say that c=λfc=\lambda f , where cc is the speed of light.
Hence c=3×108ms1c=3\times {{10}^{8}}m{{s}^{-1}} .
Now, period and frequency of a wave are related as, T=1fT=\dfrac{1}{f}
As per the given data in the question, when the period is 2×1010sec2\times {{10}^{-10}}\sec , the frequency of the light wave will be,

& f=\dfrac{1}{T} \\\ & \Rightarrow f=\dfrac{1}{(2\times {{10}^{-10}})s} \\\ & \Rightarrow f=0.5\times {{10}^{10}}{{s}^{-1}} \\\ \end{aligned}$$ **We now have the velocity and the frequency of the light wave. From equation(1) , the wavelength of the light wave is given as, $$\begin{aligned} & \lambda =\dfrac{V}{f} \\\ & \Rightarrow \lambda =\dfrac{3\times {{10}^{8}}}{0.5\times {{10}^{1O}}}m \\\ & \Rightarrow \lambda =6\times {{10}^{-2}}m=0.06m \\\ \end{aligned}$$ .** **Note:-** It is important to make sure that the units of the involved physical quantities are in the same unit system. Sometimes the wavelength is given in Angstrom or nanometers, hence before proceeding to substitute the values, make sure you’ve converted them into compatible units. Also, as the wavelength of a wave of any given velocity increases, the frequency of the wave decreases and vice-versa.