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What is the relation between amplitude of a wave and its intensity?
$\begin{align}
  & A)Amplitude\propto Intensity \\
 & B)Amplitude\propto {{(Intensity)}^{2}} \\
 & C)Amplitude\propto \sqrt{Intensity} \\
 & D){{(Amplitude)}^{2}}\propto \sqrt{Intensity} \\
\end{align}$

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Last updated date: 20th Jun 2024
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Answer
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Hint: Energy of wave is proportional to the square of the amplitude of light wave in terms of wave picture of light. The intensity of a wave is equal to the power transferred by the wave per unit area. Hence, energy is proportional to the intensity.

Formula used:
$E\propto I$, where E is the energy of the wave and I is the intensity.
$E\propto {{(Amplitude)}^{2}}$

Complete step by step answer:
For light waves, the energy of the light wave is proportional to the intensity.
$E\propto I$, where E is the energy of the wave and I is the intensity.
$E\propto {{(Amplitude)}^{2}}$…(1)
Also, the intensity of a wave is power transferred per unit area.
We know that power is energy expended per unit time.
Therefore,
$I=\dfrac{E}{At}$, where, A is the area of the wave and t is the time.
Therefore, we can say that.
$I\propto E$…(2)
From expression (1) and (2) we can say that.
$I\propto {{(Amplitude)}^{2}}$
Or
$Amplitude\propto \sqrt{Intensity}$
Hence, our answer is option C.

Additional Information:
This answer is according to the wave picture of light.
According to the particle picture the intensity can be derived from photon flux.
According to De Broglie principle, light as well as matter have dual (wave and particle) nature.
Light consists of electric and magnetic waves, and hence it is called electromagnetic wave.
In photoelectric effect, the more is the intensity, the more is the photocurrent if the energy of light is higher than the work function of the material.
The amplitude for a wave is the maximum value in its sinusoidal deviation.
Just like light electrons are also known to undergo wave phenomena like diffraction and interference.
It is usually not feasible to measure the intensity for visible light, but it is feasible to measure for radio waves.
The electric wave and magnetic wave are perpendicular to each other during the propagation of light.

Note:
Those students who try to solve this question in a hurry can make simple error in observation as we usually don’t see the expression $Amplitude\propto \sqrt{Intensity}$, and can confuse with the expression $Intensity\propto Amplitud{{e}^{2}}$. So, be alert while solving the question.