A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in a Young's double-slit experiment.
(a) Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm.
(b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide?

Explanation

Here, `d=2mm=2xx10^(-3)m`
D = 120 cm = 1.2 m
`lamda_(1)=650nm=65xx10^(-8)m`
`lamda_(2)=520nm=52xx10^(-8)m`
Suppose minimum distance required is x, where `n_(1)` and `n_(2)` order of bright fringes of lights with wavelengths `lamda_(1)` and `lamda_(2)` respectively coincide.
Hence,
`(xd)/(D)=nlamda=` constant and minimum ......(1)
`:.n_(1)lamda_(1)=n_(2)lamda_(2)`
`:.(n_(1))/(n_(2))=(lamda_(2))/(lamda_(1))`
`:.(n_(1))/(n_(2))=(520)/(650)=(4)/(5)`
Here minimum individual values of `n_(1)` and `n_(2)`, which can satisfy above condition are `n_(1)=4` and `n_(2)=5`. Hence, from equation (1),
`(xd)/(D)=4lamda_(1)=5lamda_(2)" "....(2)`
Let us take `(xd)/(D)=4lamda_(1)` [From equation (1)]
`:.x=(4lamda_(1)D)/(d)`
`=((4)(650xx10^(9))(120xx10^(-2)))/((2xx10^(-3)))`
`:.x=1.56xx10^(-3)m`
Here we may take `(xd)/(D)=5lamda_(2)` [(From equation (2)]
`:.x=(5lamda_(2)D)/(d)`
`=((5)(520xx10^(-9))(120xx10^(-2)))/((2xx10^(-3)))`
`=1.56xx10^(-3)m`