Light of wavelength 687 nm is incident on a single slit 0.75 mm wide. At what distance from the slit should a screen be placed if the second dark fringe in the diffraction pattern is to be 1.7 mm from the center of the screen?
A) 0.39 m
B) 0.93 m
C) 1.1 m
D) 1.5 m
E) 1.9 m

Question

Quizplus · Accepted Answer

The position of the dark fringes in a single-slit diffraction pattern is given by the formula $ a \sin \theta = m \lambda $, where $ a $ is the slit width, $ \lambda $ is the wavelength, $ m $ is the order of the dark fringe, and $ \theta $ is the angle. For small angles, $ \sin \theta \approx \tan \theta = \frac{y}{L} $, where $ y $ is the fringe distance from the center and $ L $ is the distance to the screen. For the second dark fringe ($ m = 2 $), $ a \frac{y}{L} = 2 \lambda $. Solving for $ L $ gives $ L = \frac{a y}{2 \lambda} $. Substituting $ a = 0.75 \times 10^{-3} $ m, $ y = 1.7 \times 10^{-3} $ m, and $ \lambda = 687 \times 10^{-9} $ m, we find $ L \approx 0.93 $ m.

Light of Wavelength 687 Nm Is Incident on a Single