Question

Two coherent sources of intensity ratio 9:4 produce interference. The intensity ratio of maxima and minima of interference pattern is:
A. 13:5
B. 5:1
C. 25:1
D. 3:2

Answer 1

Hint: We have to keep in mind the relation between intensity and amplitude. As intensity is directly proportional to the square of amplitude $I\propto {{A}^{2}}$. Maximum intensity is the sum of amplitudes of two waves and then squaring their sum i.e. ${{I}_{\max }}\propto {{\left( {{A}_{1}}+{{A}_{2}} \right)}^{2}}$ and minimum intensity is the difference of amplitudes and the squaring their difference i.e. ${{I}_{\min }}\propto {{\left( {{A}_{1}}-{{A}_{2}} \right)}^{2}}$.

Formula Used:
$I\propto {{A}^{2}}$
Where:
I is the intensity and,
A is the amplitude

Complete step by step answer:
To move further firstly, we have to know the coherent source and interference.
Coherent Source: Two sources said to be coherent if the waves associated with them have the same frequency, constant phase difference and nearly same amplitude.
Interference: Interference is the phenomenon of the superimposition (or overlapping) of waves. The waves that are superimposed must be coherent or having the same frequency and constant phase difference. As a result we see a pattern of light and dark fringes.
In our question we are given with the ratio of intensities of two waves:
$\dfrac{{{I}_{1}}}{{{I}_{2}}}=\dfrac{9}{4}$
And we know that $I\propto {{A}^{2}}$
Therefore replacing intensity with amplitude we get,
$\dfrac{{{A}_{1}}^{2}}{{{A}^{2}}_{2}}=\dfrac{9}{4}$
Taking square roots on both side
$\dfrac{{{A}_{1}}}{{{A}_{2}}}=\dfrac{3}{2}$
For maximum intensity of two coherent source,
${{I}_{\max }}\propto {{\left( {{A}_{1}}+{{A}_{2}} \right)}^{2}}$
And for minimum intensity of two source,
${{I}_{\min }}\propto {{\left( {{A}_{1}}-{{A}_{2}} \right)}^{2}}$
Here, we have $\dfrac{{{A}_{1}}}{{{A}_{2}}}=\dfrac{3}{2}$
By applying Componendo and Dividendo we get,

& \dfrac{{{A}_{1}}}{{{A}_{2}}}+1=\dfrac{3}{2}+1 \\\ & \dfrac{{{A}_{1}}+{{A}_{2}}}{{{A}_{2}}}=\dfrac{3+2}{2} \\\ & \dfrac{{{A}_{1}}+{{A}_{2}}}{{{A}_{2}}}=\dfrac{5}{2} \\\ \end{aligned}$$ Squaring both side we get $$\dfrac{{{({{A}_{1}}+{{A}_{2}})}^{2}}}{{{({{A}_{2}})}^{2}}}=\dfrac{25}{4}$$ ……….(1) And, $$\begin{aligned} & \dfrac{{{A}_{1}}}{{{A}_{2}}}-1=\dfrac{3}{2}-1 \\\ & \dfrac{{{A}_{1}}-{{A}_{2}}}{{{A}_{2}}}=\dfrac{3-2}{2} \\\ & \dfrac{{{A}_{1}}-{{A}_{2}}}{{{A}_{2}}}=\dfrac{1}{2} \\\ \end{aligned}$$ $$\dfrac{{{({{A}_{1}}-{{A}_{2}})}^{2}}}{{{({{A}_{2}})}^{2}}}=\dfrac{1}{4}$$ ………(2) Dividing equation (1) and (2) we get, $$\dfrac{{{({{A}_{1}}+{{A}_{2}})}^{2}}}{{{({{A}_{1}}-{{A}_{2}})}^{2}}}=\dfrac{25}{1}$$ Or $$\dfrac{{{I}_{\max }}}{{{I}_{\min }}}=\dfrac{25}{1}$$ Therefore the ratio of maximum intensity to the minimum intensity is 25:1. Option (C) is correct. Note: Most of the students can make mistakes in the ratio of maximum intensity to minimum intensity. They just assume that maximum intensity is the sum of intensities and minimum intensity is the difference of intensities. But in actuality, it is the amplitude which determines the result of intensities.