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In YDSE a = 2mm, D = 2m, lambda = 500 nm...

In YDSE `a = 2mm`, D = 2m, `lambda = 500` nm. Find distance of point on screen from central maxima where intensity becomes `50%` of central maxima

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`I_("max")= I_(1)= 4a^(2)" "` (at central maximum)
`I_(2)= 75% " of "I_(1)" "` (at point P)
`=(3I)/(4)= (3)/(4)xx4a^(2)= 3a^(2)`
Resultant amplitude at .P. is `A= sqrt(3)a`
If `phi` is phase difference `A^(2)= A_(1)^(2)+A_(2)^(2)+2A_(1)A_(2) cos phi`
`3Aa^(2)= A^(2)+A^(2)+2A^(2)cos phi : phi =(pi)/(3)`
Then corresponding path difference
`phi = (2pi x)/(lambda) implies x= (lambda)/(6) therefore` Path difference `x= y(d)/(D)`
`y= 4.8xx 10^(-5)m`.
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