In Young's double slit experiment, the two slits `0.15 mm` apart are illuminated by light of wavelength `450 nm`. The screen is `1.0 m` away from the slits. Find the distance of second bright fringe and second dark fringe from the central maximum. How will the fringe pattern change if the screen is moved away from the slits ?

Answer

`6 mm ; 4.5 mm`

Explanation

Here, `d = 0.15 mm = 0.15 xx 10^(-3)m`,
`lambda = 450 nm = 450 xx 10^(-9)m, D = 1m`
Distance of 2nd bright fringe from central max.
`x_(2) = 2(lambda D)/(d) = (2 xx 450 xx10^(-9) xx 1)/(0.15 xx 10^(-3))`
`= 6 xx 10^(-3)m = 6mm`
Distance of 2nd dark fringe from central max.
`x'_(2) = (2n - 1)(lambda D)/(d) = (3 xx 450 xx 10^(-9) xx 1)/(2 xx 0.15 xx 10^(-3))`
`= 4.5 xx 10^(-3) m`
`= 4.5 mm`