In Young's double slit experiment, a minimum is obtained when the phase difference of super imposing waves is

Options

A. zero
B. `(2n-1)pi`
C. `n pi`
D. `(n+1)pi`

Answer

B

Explanation

For minima at a point the phase difference between the two waves reaching the point should be an odd integral mulitple of `pi`.
As resultant amplitude of two superimposed wave is
`R=sqrt(a_(1)^(2)+a_(2)^(2)+2a_(1)a_(2)cos phi) `
and resultant intensity of two surperimposed wave is
`I=I_(1)+I_(2)+2ka_(1)a_(2)cos phi`
For minima (destructive interference), phase difference should be odd multiple of `pi`.
i.e. `phi=(2n-1)pi, " where " n=1, 2, 3, ...`
As `phi=pi,3pi,5pi, ...`
Hence, `cos phi=cos(2n-1)pi=-1`
So, `R_("min")=sqrt(a_(1)^(2)+a_(2)^(2)-2a_(1)a_(2))=sqrt((a_(1)-a_(2))^(2))`
or `R_("min")=a_(1)-a_(2)`
Also, `I=I_(1)+I_(2)-2ka_(1)a_(2)`
or `I=I_(1)+I_(2)-2a_(1)sqrt(k) a_(2)sqrt(k)`
`=I_(1)+I_(2)-2 sqrt(I_(1)I_(2))`
Hence, `I lt I_(1)+I_(2)`
Hence, when the phase difference is odd multiple of `pi`, then destructive interference is obtained i.e., `(2n-1)pi`.