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 intensity of the waves are in the ratio:
1. 16 : 9
2. 25 : 9
3. 4 : 1
4. 5 : 3

Answer 1

Correct Answer - Option 2 : 25 : 9

Concept:

Resultant Intensity:

Resultant Intensity of two waves is given as,

\(I = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}} {\rm{cos}}\theta \)  

Where, I₁ is the intensity of coherent source 1

I₂ is the intensity of coherent source 1

and θ is the Phase difference

Calculation:

We know that

\(I = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}} {\rm{cos}}\theta \)   ----(1)  

Given, the ratio of the maximum intensity to the minimum intensity = 16 : 1

Then, Maximum intensity, \({I_{{\rm{max}}}} = {\left( {\sqrt {{I_1}} + \sqrt {{I_2}} } \right)^2}\)

Minimum intensity, \({I_{{\rm{min}}}} = {\left( {\sqrt {{I_1}} - \sqrt {{I_2}} } \right)^2}\) 

\( \Rightarrow \frac{{{I_{{\rm{max}}}}}}{{{\rm{\;}}{I_{{\rm{min}}}}}} = \frac{{{{\left( {\sqrt {{I_1}} + \sqrt {{I_2}} } \right)}^2}}}{{{{\left( {\sqrt {{I_1}} - \sqrt {{I_2}} } \right)}^2}}} = \frac{{16}}{1}\) 

By simplifying

\( \Rightarrow \frac{{\sqrt {{I_1}} + \sqrt {{I_2}} }}{{\sqrt {{I_1}} - \sqrt {{I_2}} }} = \sqrt {\frac{{16}}{1}} \) 

\( \Rightarrow \frac{{\sqrt {{I_1}} + \sqrt {{I_2}} }}{{\sqrt {{I_1}} - \sqrt {{I_2}} }} = 4\)

\( \Rightarrow \sqrt {{I_1}} + \sqrt {{I_2}} = 4\left( {\sqrt {{I_1}} - \sqrt {{I_2}} } \right)\)

\( \Rightarrow \sqrt {{I_1}} + \sqrt {{I_2}} = 4\sqrt {{I_1}} - 4\sqrt {{I_2}} \)

\( \Rightarrow \sqrt {{I_2}} + 4\sqrt {{I_2}} = 4\sqrt {{I_1}} - \sqrt {{I_1}} \)

\( \Rightarrow 5\sqrt {{I_2}} = 3\sqrt {{I_1}} \)

\( \Rightarrow \frac{{\sqrt {{I_1}} }}{{\sqrt {{I_2}} }} = \frac{5}{3}\)

Squaring on both sides, we get

\(\therefore \frac{{{I_1}}}{{{I_2}}} = \frac{{25}}{9}\) 

Therefore, option (2) is the correct answer.