In Young's double silt experiment, the two slits are 0.15 mm apart. The light source has a wavelenght of 450 nm. The screen is 2 m away from the slits.
(i) Find the distance of the second bright frings and also third dark frings from the central maximum.
(ii) Find the fringe width.
(iii) How will the frings pattern change if the screen is moved away from the silis?
(iv) what will happen to the fringe width if the whole setup is immersed in water of refractive index 4/3.

Explanation

`d = 0.15 mm = 0.15 xx 10^(-3) m: D = 2 m,`
`lambda = 450 = 450 xx 10^(-9) m, n = 4//3`
(i) Equation for `n^(th)` bright fringe is, `y_(n)=n(lambdaD)/(d)`
Distance of `2^(nd)` bright fringe is, `y=2xx(450xx10^(-9)xx2)/(0.15xx10^(-3))`
`y_(2) = 12 xx 10^(-3) m = 12 mm`
Equation for `n^(th)` dark fringe is, `y_(n)=((2n-1))/(d)(lambda)/(d)`
Distance of `3^(rd)` dark fringe is, `y_(3)=(5)/(2)xx(450xx10^(-9)xx2)/(0.15xx10^(-3))`
`y_(2)=15xx10^(-3)m=15mm`
(ii) Equation for fringe width is, `beta = (lambdaD)/(d)`
Substituting, `beta = (450 xx 10^(-9) xx 2)/(0.15 xx 10^(-3))`
`beta = 6 xx 10^(-3) m = 6 mm`
(iii) The fringe width will increase as D is increased, `beta = (lambda D)/(d) (or) beta prop D)`
The fringe width will decreases as the setup is immersed in water of refractive index 4/3. `beta = (lambdaD)/(d) (or) beta prop lambda`
The wavelenght will decrease refractive index n times. Hence `beta prop lambda and beta. prop lambda.`
We know that , `lambda = (lambda)/(n)`
`(beta.)/(beta)=(lambda.)/(lambda)=(lambda//n)/(lambda)=(1)/(n)(or)beta.(beta)/(n)=(6xx10^(3))/(4//3)`
`beta = 4.5 xx 10^(-3) m = 4.5 mm`