Calculate the fringe width in a Young's double-slit experiment given wavelength, slit separation, and screen distance.

The wavelength of light is $\lambda = 600 \text{ nm}$.

$$\lambda = 600 \times 10^{-9} \text{ m}$$

The distance between the slits is $d = 0.5 \text{ mm}$.

$$d = 0.5 \times 10^{-3} \text{ m}$$

The distance of the screen from the slits is $D = 1.5 \text{ m}$.

$$D = 1.5 \text{ m}$$

The fringe width, often denoted as $\beta$ or $w$, is given by the formula:

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

$$\beta = \frac{(600 \times 10^{-9} \text{ m}) \times (1.5 \text{ m})}{(0.5 \times 10^{-3} \text{ m})}$$ $$\beta = \frac{900 \times 10^{-9} \text{ m}^2}{0.5 \times 10^{-3} \text{ m}}$$ $$\beta = \frac{900}{0.5} \times 10^{-9} \times 10^{3} \text{ m}$$ $$\beta = 1800 \times 10^{-6} \text{ m}$$ $$\beta = 1.8 \times 10^{-3} \text{ m}$$

The fringe width can also be expressed in millimeters:

$$\beta = 1.8 \text{ mm}$$

The fringe width is $\text{1.8 mm}$.