In a double slit experiment with monochromatic light, fringes are obtained on a screen placed at some distance from the plane of slits. If the screen is moved by $$5 \times 10^{-2}$$ m towards the slits, the change in fringe width is $$3 \times 10^{-3}$$ cm. If the distance between the slits is 1 mm, then the wavelength of the light will be ______ nm.

We need to find the wavelength of light in a double slit experiment from the change in fringe width when the screen is moved.

Recall the fringe width formula $$\beta = \frac{\lambda D}{d}$$ where $$\lambda$$ is wavelength, D is the distance from slits to screen, and d is the slit separation.

When the screen is moved towards the slits by $$\Delta D = 5 \times 10^{-2}$$ m, the fringe width decreases by $$\Delta\beta = 3 \times 10^{-3}$$ cm $$= 3 \times 10^{-5}$$ m. $$\Delta\beta = \frac{\lambda \cdot \Delta D}{d}$$

d = 1 mm = 10^{-3} m. $$\lambda = \frac{\Delta\beta \times d}{\Delta D} = \frac{3 \times 10^{-5} \times 10^{-3}}{5 \times 10^{-2}}$$

$$= \frac{3 \times 10^{-8}}{5 \times 10^{-2}} = 6 \times 10^{-7} \text{ m}$$

$$= 600 \text{ nm}$$

The answer is 600 nm.