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Wave Optics question

2020 · 6 Sep · Shift 1 · Q46
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Wave Optics question

2020 · 6 Sep · Shift 1 · Q46

JEE MainPhysicsWave OpticsMCQ+4 / −1
In the figure below, P and Q are two equally intense coherent sources emitting radiation of wavelength 20 m. The separation between P and Q is 5 m and the phase of P is ahead of that of Q by 90o. A, B and C are three distinct points of observation, each equidistant from the midpoint of PQ. The intensities of radiation at A, B, C will be in the ratio : JEE Main 2020 (Online) 6th September Morning Slot Physics - Wave Optics Question 99 English
  1. A
    4 : 1 : 0
  2. B
    2 : 1 : 0
  3. C
    0 : 1 : 2
  4. D
    0 : 1 : 4
View written solutionFree

Correct answer: C

  1. Resultant intensity formula

For two coherent sources of equal intensity I0I_0I0​, the resultant intensity at a point is

I=I1+I2+2I1I2cos⁡ΔϕI = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\Delta\phiI=I1​+I2​+2I1​I2​​cosΔϕ

Since the two sources are equally intense,

I1=I2=I0I_1 = I_2 = I_0I1​=I2​=I0​

so

I=2I0(1+cos⁡Δϕ)I = 2I_0(1+\cos\Delta\phi)I=2I0​(1+cosΔϕ)

where Δϕ\Delta\phiΔϕ is the total phase difference at the observation point.


  1. Given source phase difference

Source PPP is ahead of source QQQ by 90∘90^\circ90∘. Thus initial phase difference is

ϕP−ϕQ=π2\phi_P - \phi_Q = \frac{\pi}{2}ϕP​−ϕQ​=2π​

Also, wavelength is

λ=20 m\lambda = 20\,\text{m}λ=20m

and separation between sources is

PQ=5 mPQ = 5\,\text{m}PQ=5m
  1. Path difference at points AAA, BBB, CCC

From the figure (standard geometry here), the three points are equidistant from the midpoint of PQPQPQ and lie along directions such that the path differences are:

  • At AAA: Δr=+5 m\Delta r = +5\,\text{m}Δr=+5m
  • At BBB: Δr=0\Delta r = 0Δr=0
  • At CCC: Δr=−5 m\Delta r = -5\,\text{m}Δr=−5m

So the corresponding phase differences due to path are

Δϕpath=2πλΔr\Delta\phi_{\text{path}} = \frac{2\pi}{\lambda}\Delta rΔϕpath​=λ2π​Δr

Since

2πλ=2π20=π10\frac{2\pi}{\lambda} = \frac{2\pi}{20} = \frac{\pi}{10}λ2π​=202π​=10π​

for Δr=5\Delta r = 5Δr=5 m,

Δϕpath=π10⋅5=π2\Delta\phi_{\text{path}} = \frac{\pi}{10}\cdot 5 = \frac{\pi}{2}Δϕpath​=10π​⋅5=2π​

Thus,

  • At AAA: path phase difference =+π2= +\frac{\pi}{2}=+2π​
  • At BBB: path phase difference =0= 0=0
  • At CCC: path phase difference =−π2= -\frac{\pi}{2}=−2π​

  1. Total phase difference at each point

Taking total phase difference as

Δϕ=(initial phase difference)+(path phase difference)\Delta\phi = \text{(initial phase difference)} + \text{(path phase difference)}Δϕ=(initial phase difference)+(path phase difference)

we get:

At AAA

ΔϕA=π2+π2=π\Delta\phi_A = \frac{\pi}{2} + \frac{\pi}{2} = \piΔϕA​=2π​+2π​=π

Hence,

IA=2I0(1+cos⁡π)=2I0(1−1)=0I_A = 2I_0(1+\cos\pi)=2I_0(1-1)=0IA​=2I0​(1+cosπ)=2I0​(1−1)=0

At BBB

ΔϕB=π2\Delta\phi_B = \frac{\pi}{2}ΔϕB​=2π​

Hence,

IB=2I0(1+cos⁡π2)=2I0(1+0)=2I0I_B = 2I_0(1+\cos\tfrac{\pi}{2})=2I_0(1+0)=2I_0IB​=2I0​(1+cos2π​)=2I0​(1+0)=2I0​

At CCC

ΔϕC=π2−π2=0\Delta\phi_C = \frac{\pi}{2}-\frac{\pi}{2}=0ΔϕC​=2π​−2π​=0

Hence,

IC=2I0(1+cos⁡0)=2I0(1+1)=4I0I_C = 2I_0(1+\cos 0)=2I_0(1+1)=4I_0IC​=2I0​(1+cos0)=2I0​(1+1)=4I0​
  1. Intensity ratio

Therefore,

IA:IB:IC=0:2I0:4I0=0:1:2I_A : I_B : I_C = 0 : 2I_0 : 4I_0 = 0:1:2IA​:IB​:IC​=0:2I0​:4I0​=0:1:2
  1. Compare with options

This matches:

0:1:2\boxed{0:1:2}0:1:2​

So the correct option is C.


  1. Comparison with stored answer

Stored correct answer is B: 2:1:02:1:02:1:0, but the derived answer is C: 0:1:20:1:20:1:2.

Hence, I disagree with the stored answer. The likely reason is that points AAA and CCC were interchanged while interpreting the geometry. If the labels are in the opposite order, the ratio becomes 2:1:02:1:02:1:0, but for the natural left-to-right ordering with path differences +5,0,−5+5,0,-5+5,0,−5, the ratio is 0:1:20:1:20:1:2.

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