In Young's double slit experiment to obtain interference pattern, light used consist of two wavlenghts 5200Å and 6500Å. The distance between the two slits is 2 mm and distance between slit and screen si 1.2m least bright band due to wavelenght overlaps is

Options

A. `0.6xx10^(-4) m`
B. `1.56xx10^(-4) m`
C. `25.6xx10^(-4) m`
D. `15.6xx10^(-4) m`

Answer

D

Explanation

`lamda_(1)=5200A^(@),lamda_(2)=6500A^(@),d=2mm,D=1.2m,n=?x_(n)=?`
At least distance from central bright band, `m^(th)` bright band of `lamda_(1)`, coincides with `n^(th)` brigh band of `lamda_(2)` provided `lamda_(1)ltlamda_(2)`
`:.x_(m)" of "lamda_(1)=x_(n)" of "lamda_(2)`
`(D)/(d)mlamda_(1)=(D)/(d)nlamda_(2)`
`mlamda_(1)=nlamda_(2)`
`:.(m/n)=(lamda_(2))/(lamda_(1))=(5200)/(6500)=5/4`
Thus , distance of `5^(th)` bright band of `lamda_(1)` is given by, `x_(5b)=5beta_(1)`
`=5(D)/(d)lamda_(1)`
`:.x_(5b)=(5xx1.2xx5.2xx10^(-7))/(2xx10^(-3))=1.56xx10^(-3)m`