Question

Ratio of intensity of two waves is 25:36. If interference occurs, then ratio of maximum and minimum intensity should be .
A) 61 : 11
B) 5 : 6
C) 11 : 1
D) 121 : 1​

Answer 1

Given :

Ratio of intensity of two waves = 25:36

To Find :

ratio of maximum and minimum intensity

Solution :

Let the intensities $I_{1}$ and $I_{2}$  be 25x and 36x respectively.

$\frac{maximum intensity}{minimum intensity}$   =  $(\frac{\sqrt{I_{2} } + \sqrt{I_{1} } }{\sqrt{I_{2} } -\sqrt{I_{1} } })^{2}$

                            = $(\frac{\sqrt{36x } + \sqrt{25x_} }{\sqrt{36x} -\sqrt{25x } })^{2}$

                            = $(\frac{6\sqrt{x} + 5\sqrt{x} }{6\sqrt{x} - 5\sqrt{x} })^{2}$

                            = $\frac{11\sqrt{x} }{\sqrt{x} }$

                            = $\frac{11}{1}$

∴ The ratio of maximum and minimum intensity is 11:1.

The correct option is (c) 11:1

 

Answer 2

The ratio of intensities of two waves is 25 : 36.

if interference occurs , then the ratio of maximum and minimum intensity should be ..

A) 61 : 11

B) 5 : 6

C) 11 : 1

D) 121 : 1

when two waves of intensities I₁ and I₂ are interfere , then the intensity of resultant wave is given by,

$\quad\bf I=\sqrt{I_1+I_2+2\sqrt{I_1I_2}cos\theta}$

for maximum intensity, θ = 0°

and maximum intensity,

$I_{max}=\sqrt{(I_1+I_2+2\sqrt{I_1I_2})}=(\sqrt{I_1}+\sqrt{I_2})$

for minimum intensity, θ = 180°

and minimum intensity,

$I_{min}=\sqrt{I_1+I_2-2\sqrt{I_1I_2}}=(\sqrt{I_1}-\sqrt{I_2})$

now the ratio of maximum intensity to minimum intensity is..

$\frac{I_{max}}{I_{min}}=\left|\frac{(\sqrt{I_1}+\sqrt{I_2})}{(\sqrt{I_1}-\sqrt{I_2})}\right|$

= $\left|\frac{\sqrt{\frac{I_1}{I_2}}+1}{\sqrt{\frac{I_1}{I_2}}-1}\right|$

here given, $\frac{I_1}{I_2}=\frac{25}{36}$

= $\left|\frac{\sqrt{\frac{25}{36}}+1}{\sqrt{\frac{25}{36}}-1}\right|$

= $\left|\frac{5+6}{5-6}\right|$

= $\frac{11}{1}$

Therefore the ratio of maximum intensity to minimum intensity is 11 : 1. hence option (C) is correct choice.