Question

Two coherent sources of different intensities send waves that interfere. The ratio of maximum to minimum intensity is $25$. The intensity ratio of the sources is
A. $25:1$
B. $5:1$
C. $9:4$
D. $25:16$

Answer 1

Hint: First assume two different amplitudes for these two waves. Then applying intensity formula one for maximum intensity and another for minimum intensity for these two waves. The Ratio of the maximum to minimum intensity is given in the question.With the help of this we can calculate the ratio of intensity of the sources.

Complete step by step answer:
As per the given problem there are two coherent sources of different intensities that interfere with each other and produce maximum and minimum intensity. And it is given that their ratio is 25.Intensity is defined by the square of its amplitude.Let ${a_1}$ and ${a_2}$ be the amplitude of the two waves respectively.Maximum intensity of two interfered waves is calculated by sum of amplitudes of waves in the whole square.

Similarly the minimum intensity of two interfered waves is calculated by the difference of amplitude of the whole square.Mathematically we can write,
${I_{\max }} = {\left( {{a_1} + {a_2}} \right)^2} \ldots \ldots \left( 1 \right)$
$ \Rightarrow {I_{\min }} = {\left( {{a_1} - {a_2}} \right)^2} \ldots \ldots \left( 2 \right)$
Taking ratio of equation $\left( 1 \right)$ and $\left( 2 \right)$ we get,
$\dfrac{{{I_{\max }}}}{{{I_{\min }}}} = \dfrac{{{{\left( {{a_1} + {a_2}} \right)}^2}}}{{{{\left( {{a_1} - {a_2}} \right)}^2}}} \ldots \ldots \left( 3 \right)$

As given in the problem,the ratio of maximum intensity to minimum intensity is 25.Hence,
$\dfrac{{{I_{\max }}}}{{{I_{\min }}}} = \dfrac{{25}}{1} \ldots \ldots \left( 4 \right)$
Equating equation $\left( 3 \right)$ and $\left( 4 \right)$ we get,
$\dfrac{{25}}{1} = {\left( {\dfrac{{{a_1} + {a_2}}}{{{a_1} - {a_2}}}} \right)^2}$
Taking square to other side we get,
$\sqrt {\dfrac{{25}}{1}} = \left( {\dfrac{{{a_1} + {a_2}}}{{{a_1} - {a_2}}}} \right)$
$ \Rightarrow \left( {\dfrac{{{a_1} + {a_2}}}{{{a_1} - {a_2}}}} \right) = \dfrac{5}{1}$
Rearranging the above equation we get,
${a_1} + {a_2} = 5\left( {{a_1} - {a_2}} \right)$
$ \Rightarrow {a_1} + {a_2} = 5{a_1} - 5{a_2}$
Further solving this equation we get,
${a_2} + 5{a_2} = 5{a_1} - {a_1}$
$ \Rightarrow 6{a_2} = 4{a_1}$

Rearranging the above equation we get,
$\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{6}{4}$
Simplifying the fractional term we get,
$\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{3}{2}$
Squaring both the sides we get,
${\left( {\dfrac{{{a_1}}}{{{a_2}}}} \right)^2} = {\left( {\dfrac{3}{2}} \right)^2}$
$\Rightarrow \dfrac{{{a_1}^2}}{{{a_2}^2}} = \dfrac{9}{4} \ldots \ldots \left( 5 \right)$
Where,
$\dfrac{{{a_1}^2}}{{{a_2}^2}} = \dfrac{{{\operatorname{I} _1}}}{{{I_2}}} \ldots \ldots \left( 6 \right)$
Replacing equation $\left( 5 \right)$ with $\left( 6 \right)$ we get,
$\therefore \dfrac{{{I_1}}}{{{I_2}}} = \dfrac{9}{4}$

Therefore the correct option is $\left( C \right)$.

Note: Remember intensity is the square of the amplitude of the respective wave. For maximum intensity the amplitude is added because it forms a constructive wave form where the amplitude of the waveform increases while for minimum intensity the amplitude is subtracted because it forms destructive wave form where the amplitude of the waveform decreases.