Bichromatic light is used in YDSE having wavelengths `lambda_(1)=400nm` and `lambda_(2)=700nm` Find minimum order of `lambda_(1)` which overlaps with `lambda_(2)`

Explanation

Let `n_1` bright fringe of `lamda_1` overlaps with `n_2` bright fringe of `lamda_2`
Then `(n_1 lamda_1 D)/(d) = (n_2lamda_2 D)/(d)`
`n_1/n_2 = lamda_2/lamda_1 = 700/400 = 7/4`
The ratio `n_1/n_2 = 7/4` implies that 7th bright fringe of `lamda_1` will overlap with 4th of `lamda_2` . Similarly 14th of `lamda_1` will over lap with 8th of `lamda_2` and so on. So the minimum order of `lamda_1` which overlaps with `lamda_2` is 7.