In a Young's double slit experiment, the slits are `2mm` apart and are illuminated with a mixture of two wavelength `lambda_0=750nm` and `lambda=900nm`. The minimum distance from the common central bright fringe on a screen `2m` from the slits where a bright fringe from one interference pattern coincides with a bright fringe from the other is

Options

A. (a) `1.5mm`
B. (b) `3mm`
C. (c) `4.5mm`
D. (d) `6mm`

Answer

C

Explanation

From the given data, note that the fringe width `(beta_1)` for `lambda_1=900nm` is greater than fringe width `(beta_2)` for `lambda_2=750nm`. This means that at though the central maxima of the two coincide, but first maximum for `lambda_1=900nm` will be further away from the first maxima for `lambda_2=750nm`, and so on. A stage may come when this mismatch equals `beta_2`, then again maxima of `lambda_1=990nm`, will coincide with a maxima of `lambda_2=750nm` let this correspond to nth order fringe for `lambda_1`. Then it will correspond to `(n+1)^(th)` order fringe for `lambda_2`.
Therefore `(nlambda_1D)/(d)=((n+1)lambda_2D)/(d)`
`impliesnxx900xx10^-9=(n+1)750xx10^-9impliesn=5`
Minimum distance from central maxima
`=(nlambda_1D)/(d)=(5xx900xx10^-9xx2)/(2xx10^-3)`