A beam of light consisting of two wavelengths, 6500 Å and 5200 Å is used to obtain interference fringes in a Young's double slit experiment `(1 Å = 10^(-10) m).` The distance between the slits is 2.0 mm and the distance between the plane of the slits and the screen in 120 cm. (a) Find the distance of the third bright fringes on the screen from the central maximum for the wavelength 6500 Å (b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide ?

Explanation

(a) The centre of the nth bright frigne is at a distance `y = (n lambda D)/(d)` from the central maximum. For the 3rd bright fringe of 6500 Å.
`y= (3xx6500xx10^(-19)m xx 1.2m)/(2xx10^(-3)m)`
(b) suppose the mth bright fringe of 6500 Å.
Then, `(mxx6500ÅxxD)/(d) = (nxx5200 ÅxxD)/(d)`
or `(m)/(n) = (5200)/(6500) = (4)/(5).`
The minimum values of m and n that satisfy this equation are 4 and 5 respectively. The distance of the 4th bright frings of 6500 Å or the 5th bright fringe of 5200 Å from the central maximum is
`y = (4xx6500xx10^(-10) m xx 1.2 m)/(2xx10^(-3)m)`
=0.156 cm ~~ 0.16 cm.