Light of wavelength $610 \mathrm {~nm}$ is incident on a slit $0.20 \mathrm {~mm}$ wide and the diffraction pattern is viewed on a screen that is $1.5 \mathrm {~m}$ from the slit. What is the width on the screen of the central maximum?

Question

Quizplus · Accepted Answer

The width of the central maximum in a single-slit diffraction pattern is given by $W = 2L\lambda / a$, where $L$ is the distance to the screen, $\lambda$ is the wavelength of the light, and $a$ is the width of the slit. Substituting $L = 1.5 \, \text{m}$, $\lambda = 610 \times 10^{-9} \, \text{m}$, and $a = 0.20 \times 10^{-3} \, \text{m}$ into the formula gives $W = 2 \times 1.5 \times (610 \times 10^{-9}) / (0.20 \times 10^{-3}) = 0.0091 \, \text{m}$ or $0.91 \, \text{cm}$, which rounds to $0.92 \, \text{cm}$.

Light of Wavelength 610 nm610 \mathrm {~nm}610 nm Is Incident on a Slit 0.20 mm0.20 \mathrm {~mm}0.20 mm

Light of Wavelength $610 \mathrm {~nm}$ Is Incident on a Slit $0.20 \mathrm {~mm}$