Questions & Answers

The light beams of intensities in the ratio of $9:1$ are allowed to interfere. What will be the ratio of the intensities of maxima and minima?
(A) $3:1$
(B) $4:1$
(C) $25:9$
(D) $81:1$

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Hint: The intensity of luminous energy i.e. a bright light source is the power transferred by it per unit area, given that direction of propagation of the wave emitted by the luminous energy is perpendicular to the given unit area. It's S.I. unit is watt per square meter i.e. $\dfrac{W}{{{m^2}}}$. Maxima occurs when there is constructive interference of waves. Minima occur when there is destructive interference of waves.
Formula Used:
For maxima,${A_{res}} = {A_1} + {A_2}$
For minima, ${A_{res}} = {A_1} - {A_2}$
${A_{res}}$ is the resultant amplitude of the two waves of amplitude ${A_1}$ and ${A_2}$ .
For maxima,${I_{res}} = {(\sqrt {{I_1}} + \sqrt {{I_2}} )^2}$
For minima, ${I_{res}} = {(\sqrt {{I_1}} - \sqrt {{I_2}} )^2}$
${I_{res}}$ is the resultant intensity of the two waves of intensity ${I_1}$ and ${I_2}$.
$I \propto {A^2}$
$I$is the intensity and $A$ is the amplitude.

Complete Step-by-step solution:
We know that if two waves interfere i.e. tend to become a single wave then,
For maxima,
$ \Rightarrow {A_{res}} = {A_1} + {A_2}$
For minima,
$ \Rightarrow {A_{res}} = {A_1} - {A_2}$
where ${A_{res}}$ is the resultant amplitude of the two waves of amplitude ${A_1}$and${A_2}$.
Resultant amplitude means the amplitude of the wave which is going to be formed from the interfering waves which we here considered have amplitude ${A_1}$and${A_2}$.
As intensity and amplitude are related to each other by the relation that,
$ \Rightarrow I \propto {A^2}$
Where $I$is the intensity and $A$ is the amplitude.
Hence from the above three equations, we can conclude that,
For maxima,
${I_{res}} = {(\sqrt {{I_1}} + \sqrt {{I_2}} )^2}$
For minima,
 ${I_{res}} = {(\sqrt {{I_1}} - \sqrt {{I_2}} )^2}$
where ${I_{res}}$ is the resultant intensity of the two waves of intensity ${I_1}$ and${I_2}$.
Resultant intensity means the intensity of the wave which is going to be formed from the interfering waves which we here consider to have the intensities ${I_1}$ and${I_2}$.
In the question, it is given that,
$ \Rightarrow {I_1}:{I_2} = 9:1$
So let us assume that, ${I_1} = 9x$ and ${I_2} = x$
Therefore the ratio of intensities of maxima and minima can be calculated by the following equation,
$\dfrac{{{{(\sqrt {{I_1}} + \sqrt {{I_2}} )}^2}}}{{{{(\sqrt {{I_1}} - \sqrt {{I_2}} )}^2}}}$
From the values assumed above, we can substitute them in the above equation.
$\dfrac{{{{(\sqrt {9x} + \sqrt x )}^2}}}{{{{(\sqrt {9x} - \sqrt x )}^2}}}$
$ \Rightarrow \dfrac{{{{(3\sqrt x + \sqrt x )}^2}}}{{{{(3\sqrt x - \sqrt x )}^2}}}$
Upon further solving we get,
$ \Rightarrow \dfrac{{x{{(3 + 1)}^2}}}{{x{{(3 - 1)}^2}}}$
$ \Rightarrow \dfrac{{{{(4)}^2}}}{{{{(2)}^2}}} = \dfrac{4}{1}$

Therefore, the correct answer to the above question is (B) $4:1$

Note:
The formula used in the solution for resultant amplitude and intensity is formulated using the formula ${A_{res}} = \sqrt {{A_1}^2 + {A_2}^2 + 2{A_1}{A_2}\cos \theta } $ where all symbols mean the same as above mentioned and $\theta $ is the phase difference between the two interfering waves this formula is derived using Phasor mathematics which you will learn about in higher standards.