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Question: Answered & Verified by Expert
In Young's double slit experiment, the amplitudes of the two waves incident on the two slits are A and 2A. If I0 is the maximum intensity, then the intensity at a spot on the screen, where the phase difference between the two interfering waves is ϕ.
PhysicsWave OpticsKVPYKVPY 2018 (SB/SX)
Options:
  • A I0cos2ϕ/2
  • B I03sin2ϕ/2
  • C I095+4cosϕ
  • D I095+8cosϕ
Solution:
2385 Upvotes Verified Answer
The correct answer is: I095+4cosϕ

Resultant intensity when two waves with phase difference ϕ interfere is



I=I1+I2+2I1I2cosϕ



As, intensity IA2 or I=kA2, 



where k=a constant and A is amplitude.



So, I=A12+A22+2A1A2cosϕ



Here, A1=A and A2=2A



  I=A2+2A2+2A2Acosϕ



=A25+4cosϕ........(i)



Intensity is maximum I0,



when cosϕ=1



I0=A25+4×1=9A2



A2=I0/9.......(ii)



So, resultant intensity,



I=A25+4cosϕ



[From Eqs. (i) and (ii)]



I=I095+4cosϕ


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