Light of wavelength $580 \mathrm {~nm}$ is incident on a slit of width $0.30 \mathrm {~mm}$. An observing screen is placed $2.0 \mathrm {~m}$ past the slit. Find the distance on the screen of the first order dark fringe from the center of the pattern.

Question

Quizplus · Accepted Answer

The position of the first order dark fringe in a single-slit diffraction pattern is given by $ y = \frac{m \lambda L}{a} $, where $ m = 1 $ for the first dark fringe, $ \lambda = 580 \times 10^{-9} \mathrm{~m} $, $ L = 2.0 \mathrm{~m} $, and $ a = 0.30 \times 10^{-3} \mathrm{~m} $. Substituting these values gives $ y = \frac{1 \times 580 \times 10^{-9} \times 2.0}{0.30 \times 10^{-3}} = 3.87 \mathrm{~mm} $, which is approximately $ 3.9 \mathrm{~mm} $.

Light of Wavelength 580 nm580 \mathrm {~nm}580 nm Is Incident on a Slit of Width

Light of Wavelength $580 \mathrm {~nm}$ Is Incident on a Slit of Width