Question

Two coherent sources of equal intensities produce a maximum of 100 units. If the amplitude of one of the sources is reduced by 20 percent, then the maximum intensity produced will be:
$\begin{aligned} & (a)100 \\\ & (b)81 \\\ & (c)89 \\\ & (d)60 \\\ \end{aligned}$

Answer 1

For a two coherent system, the resultant or net intensity at a given point is given by the square of, sum of square root of the individual intensities. We use this to find the intensity of individual sources. Also, the intensity of a coherent light source is directly proportional to the square of its Amplitude.

Complete answer:
Let the intensity of the individual sources be given by $I$ and their amplitudes be $A$ .
Now, it has been given to us, the net intensity due to these two coherent sources at a point is 100 units. So, using the formula for net intensity which is:
$\Rightarrow {{I}_{net}}={{(\sqrt{{{I}_{1}}}+\sqrt{{{I}_{2}}})}^{2}}$
Here, we know the value of resultant intensity and initial intensities of the two sources as:
$\Rightarrow {{I}_{net}}=100$ , and
$\Rightarrow {{I}_{1}}={{I}_{2}}=I$
Putting these values in the above equation, we get:
$\begin{aligned} & \Rightarrow 100={{(\sqrt{I}+\sqrt{I})}^{2}} \\\ & \Rightarrow 100={{(2\sqrt{I})}^{2}} \\\ & \Rightarrow 100=4I \\\ & \Rightarrow I=25units \\\ \end{aligned}$
Therefore,
$\Rightarrow {{I}_{1}}={{I}_{2}}=25units$
Now, the amplitude of one of the sources has been decreased by 20 percent. Let this be the second source with intensity ${{I}_{2}}$ . Then its final amplitude (say ${{A}_{f}}$) can be given by:
$\begin{aligned} & \Rightarrow {{A}_{f}}=A-\dfrac{20}{100}A \\\ & \Rightarrow {{A}_{f}}=\dfrac{4A}{5} \\\ \end{aligned}$
Therefore, the final amplitude of the second source is $\dfrac{4A}{5}$.
Now, using the fact the intensity is directly proportional to Square of amplitude, we get:
$\begin{aligned} & \Rightarrow {{I}_{2,f}}\propto A_{f}^{2}\propto {{\left( \dfrac{4A}{5} \right)}^{2}} \\\ & \Rightarrow {{I}_{2,f}}\propto \left( \dfrac{16{{A}^{2}}}{25} \right) \\\ & \Rightarrow {{I}_{2,f}}=\dfrac{16}{25}I \\\ & \Rightarrow {{I}_{2,f}}=\dfrac{16}{25}\times 25 \\\ & \Rightarrow {{I}_{2,f}}=16units \\\ \end{aligned}$
Hence, the final resultant intensity due to the two coherent sources (say ${{I}_{f}}$) will be equal to:
$\begin{aligned} & \Rightarrow {{I}_{f}}={{(\sqrt{{{I}_{1}}}+\sqrt{{{I}_{2,f}}})}^{2}} \\\ & \Rightarrow {{I}_{f}}={{(\sqrt{25}+\sqrt{16})}^{2}} \\\ & \Rightarrow {{I}_{f}}={{(5+4)}^{2}} \\\ & \Rightarrow {{I}_{f}}=81units \\\ \end{aligned}$
Hence, the final resultant intensity due to the two coherent sources is equal to $81units$.

Hence, option (b) is the correct option.

Note:
We should always know these basic properties of Electromagnetic waves. One should be very careful in remembering these formulas as the square roots and whole squares could be interchanged mistakenly which would make our entire solution incorrect.