Answer to Question #87950 in Optics for Nicholas Legault

Question

Red light with a wavelength of 725 nm shines through a single slit, and forms a pattern on a screen placed 1.25 m away. If the width of the slit is 18.5 micrometers, find the width of the central maximum (a) in degrees and (b) in centimeters

Answer 1

The angular position $\theta_m$ of the first minimum is determined by the equation $\sin \theta_m = \lambda / d$, where $\lambda = 725 \times 10^{-9}\, \text{m}$ is the wavelength, and $d = 18.5 \times 10^{-6}\, \text{m}$ is the width of the slit. We obtain $\sin \theta_m = 0.725 / 18.5 \approx 0.0392$. In view of the smallness of this number, we have $\sin \theta_m \approx \theta_m \approx 0.0392$. The angular width of the central maximum is twice this angle: $\theta = 2 \theta_m \approx 0.0784$. The angular width in degrees is then $360\, \theta / 2 \pi \approx 4.5^\circ$. The width in centimetres is proportional to the distance to the screen $R = 125\, \text{cm}$ and is given approximately by $\theta R = 0.0784 \times 125\, \text{cm} = 9.8\, \text{cm}$.

Answer: (a) $4.5^\circ$, (b) $9.8\, \text{cm}$.