Question

For distinct interference pattern to be observed, the necessary condition is that the ratio of intensity of light emission by both the sources should be:
(A) $2:1$
(B) $1:2$
(C) $1:1$
(D) $1:4$

Answer 1

Hint To solve this question, first we have to calculate the amplitude of the resultant wave and then we have to find the intensity of the wave. On simplifying the relation obtained, the ratio can be easily calculated.

Formula Used: The formula used in the solution of this question is given as,
$\Rightarrow {R^2} = A_1^2 + A_2^2 + 2{A_1}{A_2}\cos \delta$
Here, $R$ is the resultant amplitude, ${A_1}$ is the amplitude of the first wave, ${A_2}$ is the amplitude of the second wave, and $\delta$ is the phase difference between the two waves.

Complete step by step answer
Let ${y_1}$ and ${y_2}$ be the two waves coming from the two coherent sources, which are given as,
${y_1} = {A_1}\sin \dfrac{{2\pi }}{\lambda }(vt - x)$
And, ${y_2} = {A_2}\sin \dfrac{{2\pi }}{\lambda }[(vt - x) + \delta ]$
Here, we assume that these two waves are coherent, that is they have a constant phase difference of $\delta$.
Thus, the amplitude of the resultant superimposed wave is given as,
$\Rightarrow {R^2} = A_1^2 + A_2^2 + 2{A_1}{A_2}\cos \delta$
The square of amplitude is proportional to the intensity of the wave. So, the value of ${R^2}$ measures the intensity at a point which is given by,
$\Rightarrow I = k{R^2} = k\left( {A_1^2 + A_2^2 + 2{A_1}{A_2}\cos \delta } \right)$
Now, if ${A_1} = {A_2} = A$
Then, we get,
$\Rightarrow I = k\left( {{A^2} + A_{}^2 + 2{A^2}\cos \delta } \right)$
$\Rightarrow I = k\left( {2{A^2} + 2{A^2}\cos \delta } \right)$
So, we get,
$\Rightarrow I = 2k{A^2}(1 + \cos \delta )$
Using the trigonometric relation,
$\Rightarrow \cos 2\theta = 1 + 2{\cos ^2}\theta \\\ \Rightarrow {\cos ^2}\theta = \dfrac{{1 - \cos 2\theta }}{2} \\\$
So we can write,
$\Rightarrow I = 4{A^2}{\cos ^2}(\dfrac{\delta }{2})$
This is the expression for getting a distinct interference pattern from the two light sources.
Thus, we derived that both the sources should have the same amplitude.
Hence the ratio should be,
$\Rightarrow ratio = \dfrac{{{I_1}}}{{{I_2}}} = \dfrac{{A_1^2}}{{A_2^2}} = \dfrac{{{A^2}}}{{{A^2}}} \\\ \Rightarrow ratio = 1:1 \\\$
$\therefore$ Option (C) is the correct option as it gives the correct ratio out of the given options.

Note
Here, we have assumed that both the waves have the same amplitude. However, if the waves passing through the two sources have different amplitudes, then, the interference pattern would not occur and the intensity obtained would be of a different ratio, as intensity is directly proportional to the square of the amplitude of the wave.