A beam of light consisting of wavelengths 6000 `Å` and 4500 `Å` is used in a YDSE with D = 1 m and d = 1 mm. Find the least distance from the central maxima, where bright fringes due to the two wavelengths coincide.

Explanation

`beta_(1)=(lambda_(1)D)/(d)=(6000xx10^(-10)xx1)/(10^(-3))=0.6 mm`
`beta_(2)=(lambda_(2)D)/(d)=0.45 mm`
Let `n_(1)` th maxima of `lambda_(1)` and `n_(2)` th maxima of `lambda_(2)` coincide at a position y.
Then `y = n_(1)P_(1)=n_(2)P_(2)=LCM` of `beta_(1)` and `beta_(2)`
`rArr y = LCM` of 0.6 cm and 0.45 mm
y = 1.8 mm
At this point 3rd maxima for `6000 Å` & 4th maxima for `4500 Å` coincide