In YDSE, bichromatic light of wavelengths 400 nm and 560 nm
are used. The distance between the slits is 0.1 mm and the distance between the
plane of the slits and the screen is 1m. The minimum distance between two
successive regions of complete darkness is

Options

A. `4mm`
B. `5.6mm`
C. `14mm`
D. `28mm`

Answer

D

Explanation

Let nth darke fringe of`400nm` falls on `mth`dark of 560nm
`((2n-1))/(2)beta_(1) =((2m-1))/(2))beta_(2) rArr (2n-1)(lambda_(1)D)/(d) =(2m-1)lambda_(2)(D)/(d)`
`(2n-1)/(2m-1)=(560)/(400) rArr (2n-1)/(2m-1)=7/5`
means`4^(th)`minima of `400nm` falls on `3^(rd)` minima of `560nm`
next `11^(th)`maxima of `400nm` falls on `8^(th)`minima of `560nm`
So the sepration between two sucessive minima=`x_(11)^(th)-x_(4)^(th)=7beta rArr x_(11)^(th)-x_(4)^(th)=(7lambdaD)/(d)`
`x_(11)^(th)-x_(4)^(th)=(7xx4xx10^(-7)xx1)/(10^(-4)) rArr x_(11)^(th)-x_(4)^(th)=28xx10^(-3)m`
`x_(11)^(th)-x_(4)^(th)=28nm`