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JEE Main 20205 SeptemberEvening Shift
Exam
JEE Main
Year
2020
Date
5 September
Shift
Evening Shift
Subject
Physics
Chapter
Wave Optics

JEE Main 2020 · 5 September · Evening Shift · Physics · Wave Optics

Two coherent sources of sound, S _ 1 and S _ 2 , produce sound...

Two coherent sources of sound, S1{S}_{1} and S2,{S}_{2}, produce sound waves of the same wavelength λ=1m\lambda = 1m are in phase. S1{S}_{1} and S2{S}_{2} are placed 1.5m1 . 5m apart (see fig). A listener, located at LL, directly in front of S2{S}_{2}, finds that the intensity is at a minimum when he is 2m2m away from S2{S}_{2}. The listener moves away from S1{S}_{1}, keeping the distance from S2{S}_{2} fixed. The adjacent maximum of intensity is observed when the listener is at a distance dd from S1{S}_{1}. Then dd is :

Two coherent sources of sound, S _ 1 and S _ 2 , produce sound waves of the same wavelength λ = 1m are in phase. S _ 1

Options

  1. A12m12m
  2. B5m5m
  3. C2m2m
  4. D3m3m

Correct answer

B. 5m5m

Step-by-step solution

Given,

Two sources od same wavelength, λ=1m\lambda = 1m

Distance between two sources,S1S2=1.5m{S}_{1}{S}_{2} = 1 . 5m

Distance of listener from the source S2{S}_{2}, S2L=2m{S}_{2}L = 2m

Condition for minimum intensity, at 2m2m is

Path difference, P.D.=(2n+1)λ2...(1)P . D . = ( 2n + 1 ) \dfrac{\lambda}{2} . . . ( 1 )

⇒P.D.=22+(1.5)2=(2n+1)λ2⇒ P . D . = \sqrt{{2}^{2} + ( 1 . 5{)}^{2}} = ( 2n + 1 ) \dfrac{\lambda}{2}
⇒2.5=(2n+1)λ2...(1)⇒ 2 . 5 = ( 2n + 1 ) \dfrac{\lambda}{2} . . . ( 1 )

Now, the path difference between the rays is so that the adjacent maxima is obtained,

P.D.=d−2=nλ...(2)P . D . = d - 2 = n\lambda . . . ( 2 )

The path difference between adjacent maxima and minima is λ2\dfrac{\lambda}{2}, form equation (1) and (2) we get,

(d−2)−2.5=λ2...(3)( d - 2 ) - 2 . 5 = \dfrac{\lambda}{2} . . . ( 3 )
⇒d−2=2.5+12⇒ d - 2 = 2 . 5 + \dfrac{1}{2}

⇒d=5m⇒ d = 5m

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