Waves emitted by two identical sources produces intensity of K unit at a point on screen where path difference between these waves is `lamda`. Calculate the intensity at that point on screen at which path difference is `(lamda)/(4)`

Waves emitted by two identical sources produces intensity of K unit at

1.	Waves emitted by two identical sources produces intensity of K unit at a point on screen where path difference between these waves is `lamda`. Calculate the intensity at that point on screen at which path difference is `(lamda)/(4)`
Answer» `phi_(1)=(2pidelta)/(lamda)implies(2pi)/(lamda)xxlamda=2pi` and `phi_(2)=(2pi)/(lamda)xx(lamda)/(4)=(pi)/(2)I_(1)=I_(0)+I_(0)+2sqrt(I_(0))sqrt(I_(0))cos2pi=4I_(0)` and `I_(2)=I_(0)+I_(0)+2sqrt(I_(0))sqrt((I_(0))(cos2pi)/(2))=2I_(0)` `therefore(I_(1))/(I_(2))=(4I_(0))/(2I_(0))=2impliesI_(2)=(I_(1))/(2)=(K)/(2)` unit `[becauseI_(1)=K"unit"`]

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Waves emitted by two identical sources produces intensity of K unit at a point on screen where path difference between these waves is `lamda`. Calculate the intensity at that point on screen at which path difference is `(lamda)/(4)`

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