A linear aperture whose width is 0.02 em is placed immediately in front of a lens of focal length 60 cm. The aperture is illuminated normally by a parallel beam of wavelength `5 xx 10^(-5) cm`. The distance of the first dark band of the differation pattern from the centre of the screen is .......

Options

A. `0.20` cm
B. `0.15 cm`
C. `0.10 cm`
D. `0.25 cm`

Answer

B

Explanation

For `m^(th)` order dark fringe,
`-d sin theta_(m)=m lambda`
`:. sin theta_(1)=((1)lambda)/(d)`
but `sin theta_(1)=(x_(1))/(D)`
`:.(x_(1))/(D)=(lambda)/(d)`
`:.x_(1)=(D lambda)/(d)`
`=(60xx5xx10^(-5))/(0.02)=0.15 cm`