In an interference, the intensity of two interfering waves are I and 4I respectively. They produce intensity at two points A and B with phase angles of $\dfrac{\pi }{2}$ and $\pi $ respectively. Then the difference between them is.
A. I
B. 2I
C. 4I
D. 5I

Hint: We are given the intensities of two interfering waves and the phase angles at two points. Using the equation to find total intensity after interfering we can find the net intensities at the two points. Later by finding the difference between them we get the solution.

Formula used:
${{I}_{net}}={{I}_{1}}+{{I}_{2}}+2\sqrt{{{I}_{1}}{{I}_{2}}}\cos \phi $

Complete step-by-step answer:
In the question it is said that two waves interfere and the intensities of these two waves are given to us.
Let ‘${{I}_{1}}$’ be the intensity of the first wave. Then we are given that,
${{I}_{1}}=I$
Let ‘${{I}_{2}}$’ be the intensity of the second wave. Then we have,
${{I}_{2}}=4I$
It is said that this wave produces intensities at two points A and B and the phase angle at these points is given to us.
Let ‘${{\phi }_{A}}$’ and ‘\[{{\phi }_{B}}\]’ be the phase angle at the points A and B respectively.
It is given to us that,
${{\phi }_{A}}=\dfrac{\pi }{2}$
${{\phi }_{B}}=\dfrac{\pi }{2}$
We need to find the difference in intensities at the points A and B.
We know that the net intensity after interference is given by the formula,
${{I}_{net}}={{I}_{1}}+{{I}_{2}}+2\sqrt{{{I}_{1}}{{I}_{2}}}\cos \phi $
First let us find the net intensity at the point A.
It can be given as,
${{\left( {{I}_{net}} \right)}_{A}}={{I}_{1}}+{{I}_{2}}+2\sqrt{{{I}_{1}}{{I}_{2}}}\cos {{\phi }_{A}}$
Now we can substitute the known values in the above equation. Thus we get,
$\Rightarrow {{\left( {{I}_{net}} \right)}_{A}}=I+4I+2\sqrt{I\times 4I}\cos \left( \dfrac{\pi }{2} \right)$
Since $\cos \left( \dfrac{\pi }{2}=0 \right)$, we get
$\Rightarrow {{\left( {{I}_{net}} \right)}_{A}}=5I$
Now let us find the net intensity at point B.
We know that,
$\Rightarrow {{\left( {{I}_{net}} \right)}_{B}}={{I}_{1}}+{{I}_{2}}+2\sqrt{{{I}_{1}}{{I}_{2}}}\cos {{\phi }_{B}}$
By substituting the values we already know, we get
$\Rightarrow {{\left( {{I}_{net}} \right)}_{B}}=I+4I+2\sqrt{I\times 4I}\cos \left( \pi \right)$
Since $\cos \left( \pi \right)=-1$, by solving the above equation we get
$\Rightarrow {{\left( {{I}_{net}} \right)}_{B}}=5I-2\sqrt{4{{I}^{2}}}$
$\Rightarrow {{\left( {{I}_{net}} \right)}_{B}}=5I-\left( 2\times 2I \right)$
$\Rightarrow {{\left( {{I}_{net}} \right)}_{B}}=5I-4I$
$\Rightarrow {{\left( {{I}_{net}} \right)}_{B}}=I$
To find the difference between net intensities at points A and B,
${{\left( {{I}_{net}} \right)}_{A}}-{{\left( {{I}_{net}} \right)}_{B}}$
By substituting their calculated values, we get
$\Rightarrow 5I-I$
$\Rightarrow 4I$
Therefore the difference between the intensities at points A and B is 4I.

So, the correct answer is “Option C”.

Note: Interference of waves occurs when two waves traveling in the same medium at the same time meet each other. Due to interference there is an effect on the particles of the medium which causes the medium to change its shape. During interference the net amplitude of the two waves also increases.
There is constructive interference and destructive interference. Constructive interference occurs anywhere in the medium when the two interfering waves are displaced in the same direction. Whereas destructive interference takes place anywhere when the interfering waves are displaced in the opposite direction.