In Young's double slit experiment the two slits are $$0.6$$ mm distance apart. Interference pattern is observed on a screen at a distance $$80$$ cm from the slits. The first dark fringe is observed on the screen directly opposite to one of the slits. The wavelength of light will be ______ nm.

In Young's double slit experiment, the slit separation is 0.6 mm, the screen distance is 80 cm, and the first dark fringe appears directly opposite one of the slits.

The central bright fringe is at the midpoint of the two slits. If the first dark fringe is directly opposite one slit, its distance from the central maximum is:

$$y = \frac{d}{2} = \frac{0.6}{2} = 0.3 \text{ mm}$$

The position of the first dark fringe (m = 0) in YDSE is:

$$y = \frac{(2m+1)\lambda D}{2d} = \frac{\lambda D}{2d}$$

Equating these expressions yields

$$\frac{d}{2} = \frac{\lambda D}{2d}$$

This simplifies to

$$d^2 = \lambda D$$

Therefore, the wavelength is given by

$$\lambda = \frac{d^2}{D} = \frac{(0.6 \times 10^{-3})^2}{0.8}$$

Evaluating this gives

$$\lambda = \frac{0.36 \times 10^{-6}}{0.8} = 0.45 \times 10^{-6} \text{ m} = 450 \text{ nm}$$

The wavelength of light is 450 nm.