In a double slit interference experiment, the fringe width obtained with a light of wavelength 5900 Å was 1.2 mm for parallel narrow slits placed 2 mm apart. In this arrangement. If the slit separation is increased by one-and-half times the previous value, then the fringe width is

Options

A. 0.9 mm
B. 0.8 mm
C. 1.8 mm
D. 1.6 mm

Answer

B

Explanation

Given, `beta_(1)=1.2 mm`
`(d_(2))/(d_(1))=1(1)/(2)=1.5`
By Young's double slit interference experiment,
`beta=(lambdaD)/(d)`
So, `beta prop (1)/(d) rArr (beta_(1))/(beta_(2))=(1//d_(1))/(1//d_(2)) rArr (beta_(1))/(beta_(2))=(d_(2))/(d_(1))=1.5`
`therefore (1.2)/(beta_(2))=1.5 rArr beta_(2)=(1.2)/(1.5)=(4)/(5)=0.8 mm`