Question

Helium-neon laser light, with a wavelength of 732.8 nm, pass through a single slit with a width of 43 µm, on a screen 4 m from the slit, find the size of the central maximum

Answer 1

The condition for destructive interference in a single-slit diffraction:

$$d\text{ sin}\theta=m\lambda.$$

On the other hand, the half-width of the central maximum divided by the length to the source is the sine:

$$\text{sin}\theta=\frac{y}{L}.$$

Therefore, the width of the central maximum is

$$w=2y=2L\text{ sin}\theta=2L\frac{m\lambda}{d}=\\\space\\=2\cdot4\frac{1\cdot732.8\cdot10^{-9}}{43\cdot10^{-6}}=0.13\text{ m}.$$