In Young's double slit experiment using monochromatic light of wavelength `lambda`, the intensity of light at a point on the screen where path diff. is `lambda` is `K` units. Find the intensity of light at a point where path difference is `lambda//3`.

Explanation

Let `I_(1) and I_(2)` be the intensity of the two light of the light waves , their reslutanat intensites can be obtained as :
`I^(*) =i_(1)+I_(2) +2 sqrt(I_(1)I_(2) )cos phi`
where ,
`phi=` Phase difference between the two Waves
`I_(1)I_(2)`
`therefore I_(1)=I_(1)+I_(2)+sqrt(I_(1)I_(2)) cos phi`
`=2 I_(2) +2I_(1) cos phi `
Phase difference `=(2pi)/(lamda)xx` path diofference
since path difference `= lamda ,`
phase diffence ,`phi = 2pi`
` therefore I^(') =2I_(1)+2l_(1)=4l_(1)`
Given
`I^(')=K `
`therefore I_(4) =(K) /(4)`
when path difference `(lamda)/(3)`
phase difference `phi =(2pi)/(3)`
Hence resultanat intensity `l_(R)^(*) = l_(1)+l_(1)+2sqrt(l_(1) l_(1)) cos (2pi)/(3)`
`=2 l_(1) +2l_(1) (-(1)/(2))=l_(1)`
using equation (1) ,we are can write :
`l_(R) =L_(1) =(K)/(4)`
Hence , the intensity of light at a point where the path difference is `(lamda)/(3) ` is `(K) /(4)` units .