Question

Two interfering waves have intensities in the ratio 9 : 1. Then, the ratio of maximum to minimum amplitude is :

Answer 1

Hii mate,

◆ Answer-

Imax : Imin = 4 : 1

◆ Explaination-

# Given-

I1/I2 = 9/1

# Solution-

Intensity of the wave is directly proportional to square of amplitude.

I ∝ a^2

I1/I2 = (a1/a2)^2

9/1 = (a1/a2)^2

a1/a2 = 3

a1 = 3a2

Ratio of maximum to minimum intensity by -

Imax / Imin = (a1+a2)^2 / (a1-a2)^2

Imax / Imin = (3a2+a2)^2 / (3a2-a2)^2

Imax / Imin = 4^2 / 2^2

Imax / Imin = 16 / 4

Imax / Imin = 4

Therefore, ratio of maximum to minimum intensity is 4:1.

Hope this helps you..

Answer 2

Answer :

Let the amplitude of the waves be a$a_{1}$ and $a_{2}$ and the intensities $I_{1} and  I_{2}$ .

The ratio of intensities is 9:1.

by the formula,

$\frac{I_{1} }{I_{2} }$ = ( $\frac{a_{1} }{a_{2} }$)^2

∴ $\frac{a_{1} }{a_{2} }$ = 3:1

In interference the maximum and minimum resultant amplitudes are ($(a_{1} +a_{2}) and  (a_{1}-a_{2}  )$

∴

        maximum intensity $\frac{I_{max} }{I_{min} } = \frac{K(a_{1} +a_{2})^{2}   }{K(a_{1}-a_{2})^{2}  }$

but from the above we get $a_{1}  = 3:1 a_{2}$

∴ $\frac{I_{max} }{I_{min} } = \frac{(\frac{3}{1}a_{2}  + a_{2})^{2}   }{(\frac{3}{1} a_{2} - a_{2})^{2}   }$

= $\frac{(4a_{2 })^{2}  }{(2a_{2})^{2}  }$

= $\frac{16a_{2} }{4a_{2} }$

= 4:1

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(heya dearie pls ignore the A cap on putting formula it has been putted)

hope it helps uh :))

thanks :))