In Young's double slit experiment, the fringe width is ` 1 xx 10^(-4) m` if the distance between the slit and screen is doubled and the distance between the two slit is reduced to half and wavelength is changed from `6.4 xx 10^(7) m` to `4.0 xx 10^(-7)m` , the value of new fringe width will be

Options

A. `2.5xx10^(-4)m`
B. `2.0xx10^(-4)m`
C. `1xx10^(-4)m`
D. `0.5xx10^(-4)m`

Answer

A

Explanation

Fringe width `beta=(Dlambda)/d`
According to question `beta_(2)/beta_(1)=(lambda_(2)D_(2)d_(1))/(lambda_(1)D_(!)d_(2))`
Hence `D_(2)=2D_(1)`
`d_(2)=d_(1)/2`
`rArrbeta_(2)/beta_(1)=(lambda_(2)(2D_(1))xxd_(1))/(lambda_(1)(D_(1))xx(d_(1)/2))=4lambda_(2)/lambda_(1)=(4xx4xx10^(-7))/(64xx10^(-7))`
`rArrbeta_(2)=(4xx4xx10^(-7)xx1xx10^(-4))/(64xx10^(-7))=2.5xx10^(-4)m`