Question

Two waves having their intensities in the ratio $9:1$ produce interference. In the interference pattern, the ratio of maximum to minimum intensity is equal to
A. $2:1$
B. $9:1$
C. $3:1$
D. $4:1$

Answer 1

The two waves having intensities in a certain ratio produce an interference pattern. That means, there are both maximum and minimum intensities that exist. Maximum intensity is the whole square of the sum of the amplitudes of two waves and minimum intensity is the whole square of the difference of the amplitude of two waves. The amplitude is equal to the square root of the intensity. Hence by simplifying we can use the formula for the ratio of their minimum and maximum intensities.

Complete step by step solution:
Let the intensities of the two waves be ${I_1}$ ​ and ${I_2}$​. To proceed further, firstly, we should know the coherent source and interference.
A coherent Source is defined as the two sources that are coherent if the waves associated with them have the same frequency, constant phase difference, and almost the same amplitude.
Interference is the phenomenon of the overlapping of waves. The waves that are superimposed need to be coherent or have the same frequency and constant phase difference. Hence we see a pattern of light and dark fringes.
Given: In our question, we are given the ratio of intensities of two waves ${I_1}:{I_2} = 9:1$
The ratio of maximum and minimum intensities
$\dfrac{{{I_{\max }}}}{{{I_{\min }}}} = {\left( {\dfrac{{{A_1} + {A_2}}}{{{A_1} - {A_2}}}} \right)^2}$
The amplitude, ${A_1} = \sqrt {{I_1}}$ and ${A_2} = \sqrt {{I_2}}$
$\Rightarrow \dfrac{{{I_{\max }}}}{{{I_{\min }}}} = {\left( {\dfrac{{\sqrt {{I_1}} + \sqrt {{I_2}} }}{{\sqrt {{I_1}} - \sqrt {{I_2}} }}} \right)^2}$
$\Rightarrow \dfrac{{{I_{\max }}}}{{{I_{\min }}}} = {\left( {\dfrac{{\dfrac{{\sqrt {{I_1}} }}{{\sqrt {{I_2}} }} + 1}}{{\dfrac{{\sqrt {{I_1}} }}{{\sqrt {{I_2}} }} - 1}}} \right)^2}$
$\Rightarrow \dfrac{{{I_{\max }}}}{{{I_{\min }}}} = {\left( {\dfrac{{\sqrt 9 + 1}}{{\sqrt 9 - 1}}} \right)^2} = {\left( {\dfrac{{3 + 1}}{{3 - 1}}} \right)^2}$

$\Rightarrow \dfrac{{{I_{\max }}}}{{{I_{\min }}}} = \left( {\dfrac{4}{2}} \right) = 2:1$
The answer is option A.

Note:
The expression used in the solution for resultant amplitude and intensity is derived using the formula ${A_{res}} = \sqrt {{A_1}^2 + {A_2}^2 + 2{A_1}{A_2}\cos \theta }$and $\theta$ is the phase difference between the two interfering waves. The intensity of luminous energy is the power transferred by it per unit area, given that direction of movement of the wave discharged by the luminous energy is perpendicular to the given unit area. It's S.I. unit is watt per square meter. Maxima happens when there is constructive interference of waves. Minima happens when there is destructive interference of waves.