Two slits separated by 0 . 5 mm are illuminated by light of wavelength 500 nm . The screen is at a distance of 120 cm from the slits . The phase difference between the interfering waves at a point 3 mm on the screen from the central bright fringe is . . . . .

Options

A. ` 5 pi`
B. `pi`
C. `3 pi`
D. `7 pi`

Answer

A

Explanation

Given,
`d = 0 . 5 xx 10^(-9) m`
`lambda = 500 xx 10^(-9) m`
`D = 120 m`
`= 120 xx 10^(-2) m `
The fringe width
`beta = (lambda d)/(D)`
` = (120 xx 10^(-2) xx 0 . 5 xx 10^(-9))/(500xx 10^(-9))`
` = 12 xx 10^(-4) m`
Number of fringe in ` 3 xx 10^(-3)` m distance on screen
` = (3 xx 10^(-3))/(12 xx 10^(-4))`
= 2 . 5
= 3 rd fringe
The phase difference
` 2 pi + 2pi + pi`
`= 5 pi`