Through a narrow slit of width 2 mm, diffraction pattern is formed on a screen kept at a distance 2 m from the slit. The wavelength of the light used is 6330 A and falls normal to the slit and screen. Then, the distance between the two minima on either side of the central maximum is

Options

A. 12.7 mm
B. 1.27 mm
C. 2.532 mm
D. 25.3 mm

Answer

b

Explanation

Width of the central maxima is given by,
`w=(2 lambdaD)/a`
`(2xx6330xx10^(-10)xx2)/(2xx10^(-3))=1.27xx10^(-3)m`
`therefore w = 1.27 mm`