In Young's double slit experiment , the two slits 0.20 mm apart are illuminated by monochromatic light of wavelength 600 nm . The screen 1.0 m away from the slits .
(a) Find the distance of the second (i) bright fringe , (ii) dark fringe from the central maximum .
(b) How will the fringe pattern change if the screen is moved away from the slits ?

Explanation

Here `d = 0.5 mm=0.15xx10^(-3)m`,
`=15xx10^(-5)`
`D=1.0 m`
(a) `(i)` Distance of the second bright fringe
`x_(2)=(2lambdaD)/(d) , [therefore x_(n)=(nlambdaD)/(d)]`
`=(2xx4.5xx10^(-7)xx1.0)/(15xx10^(-5))=(2xx4.5)/(15)xx10^(-2)`
`=0.6xx10^(-2)m=6mm`
(ii) Distance of the second dark fringe
`x_(2)=(3lambdaD)/(2d) , [therefore x_(n)=((2n-1)lambdaD)/(2d)]`
`=(2xx4.5xx10^(-7)xx1.0)/(2xx15xx10^(-5))=(3xx4.5)/(30)xx10^(-2)=0.45xx10^(-2) m=4.5mm`
(b) Since frings width `beta=(lambdaD)/(d)`
When screen is moved away. `D` increases, Therefore width of the fringes increases but the angular separation `(lambda//d)` remains the same.