Two slits are 1 mm apart and the screen is located 1 m away from the slits. A light of wavelength 500 nm is used. The width of each slit to obtain 10 maxima of the double slit pattern within the central maximum of the single slit pattern is _____ $$\times 10^{-4}$$ m.

Given: slit separation $$d = 1 \text{ mm} = 10^{-3} \text{ m}$$, screen distance $$D = 1 \text{ m}$$, wavelength $$\lambda = 500 \text{ nm} = 5 \times 10^{-7} \text{ m}$$.

Width of central maximum of single-slit diffraction pattern:

The angular half-width of the central maximum is $$\theta_0 = \frac{\lambda}{w}$$. The linear half-width on the screen is $$y_0 = \frac{D\lambda}{w}$$. The full width of the central maximum is $$2y_0 = \frac{2D\lambda}{w}$$.

Fringe width of double-slit interference pattern:

The fringe spacing (distance between consecutive maxima) is $$\beta = \frac{D\lambda}{d}$$.

Number of interference maxima within the central diffraction maximum:

The number of double-slit fringes that fit within the central diffraction maximum is:

$$N = \frac{\text{Width of central maximum}}{\text{Fringe width}} = \frac{2D\lambda / w}{D\lambda / d} = \frac{2d}{w}$$

We are told $$N = 10$$:

$$\frac{2d}{w} = 10$$

$$w = \frac{2d}{10} = \frac{d}{5}$$

$$w = \frac{10^{-3}}{5} = 2 \times 10^{-4} \text{ m}$$

The answer is $$2$$.