A beam of light consisting of two wavelengths 5000 Å and 6000 Å is used to obtain interference fringes in Young's double slit experiment. The least distance from the central maxima where the bright fringes due to both wavelengths coincide, will be:

(If separation between slits = 1 mm & separation between slits & screen is 1 m)

3 mm

The bright fringes due to both wavelengths coincide then $y_1 = y_2$

Let $m^{th}$ bright fringe of wavelength $\lambda_1$ coicides with $n^{th}$ bright fringe of wavenegth $\lambda_2$

$ y_1 = \frac{m\lambda_1 D}{d} , y_2 = \frac{n\lambda_2 D}{d}$

$ \Rightarrow m\lambda_1 = n\lambda_2$

$\Rightarrow m:n = 6:5$

$\Rightarrow \text{ Minimum value of m is 6}$

$ \Rightarrow y = \frac{6\times 5\times 10^{-7}\times 1}{10^{-3}} = 3mm$