Question

Two coherent sources of equal intensities produces, intensity maximum of $100$ units. If the amplitude of one of the sources is reduced by $20\%.$ The new maximum intensity produced will be-

• A. $100$
• B. $81$
• C. $89$
• D. $60$

Answer 1

The correct option is B $81$

Let us consider the two coherent sources of intensities $I_1\text{and}{I}_2$ respectively.

Initially let, $I_1=I_2=I$

As we know,

$I_{max}={(\sqrt{I_1}+\sqrt{I_2})}^2\dots \dots \dots \left(1\right)$

$100={(\sqrt{I}+\sqrt{I})}^2={\left(2\sqrt{I}\right)}^2=4I$

$\Rightarrow I=25\text{units}$

When amplitude of one of the source is reduced by $20\%$

Now, let

$I_1=I\text{and}{I}_2\propto {\left(A_2\right)}^2$

And $A_2=A-\frac{20}{100}A=\left(\frac{4A}{5}\right)$

$\Rightarrow I_2\propto {\left(A_2\right)}^2\propto {\left(\frac{4A}{5}\right)}^2$

$\Rightarrow I_2\propto \frac{16}{25}{A}^2$

$\therefore I_2=\frac{16}{25}I\{\because I\propto (\text{Amplitude}{)}^2\}$

From (1), putting the values of $I_1$ and $I_2$ we get,

$I_{max}^{\prime }={\left(\sqrt{I+}\sqrt{\frac{16I}{25}}\right)}^2={\left(\sqrt{I+}\frac{4}{5}\sqrt{I}\right)}^2=I{\left(\frac{9}{5}\right)}^2$

$\therefore I_{max}^{\prime }=\frac{81}{25}I=\frac{81}{25}\times 25=81\text{units}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.