(a) In Young's double slit experiment, derive the condition for (i) constructive interference and (ii) destructive interference at a point on the screen.
(b) A beam of light consisting of two wavelengths, 800 nm and 600 nm is used to obtain the interference frings in a Young's double slit experiment on a screen placed 1.4 m away. If the two slits are separated by 0.28 mm, calculate the least distance from the central bright maximum where the bright fringes of the two wavelengths coincide.

Explanation

(a) Young's double slit experiment : Consider two narrow rectangular slits `S_(1) and S_(2)` placed perpendicular to the plane of paper. Slit S is placed on the perpendicular bisector on `S_(1)S_(2)` and illuminated with monochromatic light.
The slits are separated by a small distance d. A screen is placed at a distance D from `S_(1),S_(2).` Consider a point P on the screen at distance x from O. The path difference between the waves reaching P from `S_(1) and S_(2)` is :
`P=S_(2)P-S_(1)P`
Draw `S_(1)N` perpendicular to `S_(2)P.` Then,
`P=S_(2)P-S_(1)P=S_(2)P-NP=S_(2)N`
From right-angled `DeltaS_(1)S_(2)N, (S_(2)N)/(S_(2)S_(1))=sin theta`
`P=S_(2)N=S_(2)S_(1) sin theta= d sin theta`

From `Delta COP," when "theta` is small
`sin theta approx thetaapproxtan" "theta=(x)/(D) therefore P=(xd)/(D)`
For constructive interference,
`(xd)/(D)=n lamda, n=0, 1, 2, 3,....`
When n=0 `x_(n)=0,` central bright is formed at O.
For destructive interference,
`(xd)/(D)=(2n+1)(lamda)/(2) or x_(n)=(2n+1)(lamdaD)/(2D)=(1)/(2)(lamdaD)/(d),(3)/(2)(lamda D)/(d),(5)/(2)(lamda D)/(d),...`
Thus, alternate bright and dark fringers are formed on the screen.
(b) Given
`{:(lamda_(1)=800 nm=800xx10^(-9)m),(lamda_(1)=800 nm =800 xx10^(-9)m),(lamda_(2)=600 nm =600xx10^(-9)m):}`
D=1.4m
`d=0.28 mm =0.28 xx10^(-3)m`
Let `n_(1)^(th)` maximum corresponds to `lamda_(1)` coincides with `n_(2)^(th)` maximum corresponds `lamda_(2).` Then
`n_(1)(lamda_(1)D)/(d)=n_(2)(lamda_(2)D)/(d)or, (n_(1))/(n_(2))=(n_(2))/(n_(1))=(600)/(800)=(3)/(4)`
The minimum integral value of `n_(1)` is 3 and of `n_(2)` is 4. Therefore, the minimum value of y is
`y" mix"=(lamda_(1)D)/(d)=(3xx800xx10^(-9)xx1.4)/(0.28xx10^(-3))rArry" mix "=12 mm`
Position of first bright fringe.
`lamda_(A)=800 nm =800xx10^(-9)m =8xx10^(-7) m.`
`X_(A)=lamda(D)/(d)`
`=8xx10^(-7)`
`d=0.28 mm, =0.28xx10^(-3)m.`
`D=1.4m`.
`X_(A)=lamda(D)/(d)=(8xx10^(-7)xx1.4)/(0.28xx10^(-3))=4xx10^(-4)xx10=4xx10^(-3)M.`
`"For wavelength "lamda_(B)=600 nm=600xx10^(-9)m.`
`6xx10^(-7)` M. Position of first bright fringe is
`X_(B)=lamda(D)/(d)=(6xx10^(-7)xx1.4)/(0.28xx10^(-3))=3xx10^(-7)xx10^(-3)xx10=3xx10^(-3)m.`
The least distance from the central maximum, where the bright fringes coincides.
`=X_(A)-X_(B)=4xx10^(-3)-3xx10^(-3)=1xx10^(-3)m=1 nm.`