A Fraunhofer diffraction pattern is produced from a light source of wavelength 671 nm.The light goes through a single slit and onto a screen 1.9 m away.The first dark fringe is 6.1 mm from the central bright fringe.What is the slit width? (1 nm = 10−⁹ m) A) 0.42 mm B) 0.21 mm C) 0.07 mm D) 0.31 mm E) 0.034 mm

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Quizplus · Accepted Answer

The position of the first dark fringe in a single-slit diffraction pattern is given by the formula $ a \sin \theta = m \lambda $, where $ a $ is the slit width, $ \theta $ is the angle of diffraction, $ m $ is the order of the dark fringe (1 for the first dark fringe), and $ \lambda $ is the wavelength. For small angles, $ \sin \theta \approx \tan \theta = \frac{y}{L} $, where $ y $ is the distance from the central maximum to the first dark fringe, and $ L $ is the distance from the slit to the screen. Substituting the given values: $ a \approx \frac{m \lambda L}{y} = \frac{1 \times 671 \times 10^{-9} \times 1.9}{6.1 \times 10^{-3}} \approx 0.21 $ mm.

A Fraunhofer Diffraction Pattern Is Produced from a Light Source