Question

Two coherent sources produce waves of different intensities which interfere. After interference, the ratio of the maximum intensity to the minimum intensity is $16$. The intensities of the waves are in the ratio:
(A) $4:1$
(B) $25:9$
(C) $6:9$
(D) $5:3$

Answer 1

Hint
To solve this question, we have to obtain the ratio of amplitudes of the waves from the given ratio. From there, we can easily obtain the ratio of the intensities of the waves.
$I = k{A^2}$ , where $I$ is the intensity of a wave having amplitude of $A$, and $k$ is a constant.

Complete step by step answer
Let ${I_1}$ and ${I_2}$ be the intensities of the two waves.
The given ratio is $\dfrac{{{I_{\max }}}}{{{I_{\min }}}} = 16$ (1)
We know that the intensity is related to the amplitude by the relation
$I = k{A^2}$
So, the equation (1) can be written as
$\Rightarrow \dfrac{{k{A_{\max }}^2}}{{k{A_{\min }}^2}} = 16$
Cancelling $k$ we have
$\Rightarrow \dfrac{{{A_{\max }}^2}}{{{A_{\min }}^2}} = 16$
$\Rightarrow {\left( {\dfrac{{{A_{\max }}}}{{{A_{\min }}}}} \right)^2} = 16$
Taking square root both sides, we get
$\Rightarrow \dfrac{{{A_{\max }}}}{{{A_{\min }}}} = 4$ (2)
We know that the maximum amplitude in interference is equal to the sum of the amplitudes of the waves. Also, the minimum amplitude in interference is equal to the difference of the amplitudes of the waves.
So, we have
$\Rightarrow {A_{\max }} = {A_1} + {A_2}$, and
$\Rightarrow {A_{\min }} = {A_1} - {A_2}$
(${A_1}$ and ${A_2}$ are the amplitudes of the two waves)
Thus, (2) can be written as
$\Rightarrow \dfrac{{{A_1} + {A_2}}}{{{A_1} - {A_2}}} = 4$
Applying componendo and dividendo rule, we have
$\Rightarrow \dfrac{{({A_1} + {A_2}) + ({A_1} - {A_2})}}{{({A_1} + {A_2}) - ({A_1} - {A_2})}} = \dfrac{{4 + 1}}{{4 - 1}}$
$\Rightarrow \dfrac{{2{A_1}}}{{2{A_2}}} = \dfrac{5}{3}$
Cancelling $2$ we get
$\Rightarrow \dfrac{{{A_1}}}{{{A_2}}} = \dfrac{5}{3}$
Taking square on both the sides
$\Rightarrow {\left( {\dfrac{{{A_1}}}{{{A_2}}}} \right)^2} = {\left( {\dfrac{5}{3}} \right)^2}$
$\Rightarrow \dfrac{{{A_1}^2}}{{{A_2}^2}} = \dfrac{{25}}{9}$
Multiplying and dividing the LHS by$k$
$\Rightarrow \dfrac{{k{A_1}^2}}{{k{A_2}^2}} = \dfrac{{25}}{9}$
Substituting ${I_1} = k{A_1}^2$and ${I_2} = k{A_2}^2$
$\Rightarrow \dfrac{{{I_1}}}{{{I_2}}} = \dfrac{{25}}{9}$
So, the ratio of the intensities of the waves is $25:9$

Hence, the correct answer is option (B), $25:9$

Note
Don’t forget to take the squares and the square roots. It is a common mistake to confuse the intensity with the amplitude and vice-versa. As intensity is proportional to the square of the amplitude, and is not linearly proportional, so the square has to be taken for the conversion of the intensity to the amplitude. So care must be taken about this point.