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. 1.5 mm
B. 3 mm
C. 4.5 mm
D. 6 mm

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 750 . λ 2 = nm This means that although 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)` = 750 nm and so on. A stage may come when this mismatch equals β2 , then again maxima of 900 , λ1 = nm will coincide with a maxima of 750 , λ 2 = nm let this corresponds to n th order fringe for `lambda_(1)` . Then it will correspond to th (n + )th order fringe for `lambda_(2)` .
Therefore, `(nlambda_(1)D)/(d)= ((n+1)lambda_(2)D)/(d) implies n xx 900 xx 10^(-9)= (n+1)750 xx 10^(-9) implies n= 5`
Minimum distance from Central maxima
`= (n lambda_(1)D)/(d)= ( 5xx 900 xx 10^(-9) xx 2)/(2 xx 10^(-3))= 45 xx 10^(-4) m= 4.5 mm`