Q: In YDSE, bi-chromatic light of wavelength 400 nm and 560 nm are used. The distance between the slits is 0.1 mm and the distance between the plane of the slits and the screen is 1 m. The minimum distance between two successive regions of complete dark is:

Sol: Let nth minimum of 400 nm coincides with mth minima of 560 nm, then

$\large \frac{(2n-1)400}{2} = \frac{(2m-1)560}{2} $

$\large \frac{2n-1}{2m-1} = \frac{7}{5} = \frac{14}{10} = …$

i.e., 4th minima of 400 nm coincides with 3rd minima of 560 nm. Location of this minima is,

$\large y_1 = \frac{(2 \times 4-1)400 \times 10^{-9}\times 1}{2 \times 0.1 \times 10^{-3}}$

y_{1} = 14 mm

Next 11th minima of 400 nm will coincide with 8th minima of 560 nm.

Location of this minima is,

$\large y_2 = \frac{(2 \times 11 -1)400 \times 10^{-9}\times 1}{2 \times 0.1 \times 10^{-3}}$

y_{2} = 42 mm

∴ Required distance , y_{2} – y_{1} = 28 mm