A single slit,1400 nm wide,forms a diffraction pattern when illuminated by monochromatic light of 490 nm wavelength.The largest angle from the central maximum at which the intensity is zero is closest to:
A)44°
B)41°
C)38°
D)35°
E)32°

Question

Quizplus · Accepted Answer

The angle for the first minimum in a single-slit diffraction pattern is given by the equation $ a \sin \theta = m \lambda $, where $ a $ is the slit width, $ \lambda $ is the wavelength, and $ m $ is the order of the minimum. For the first minimum, $ m = 1 $. Substituting the given values, $ 1400 \sin \theta = 490 $, solving for $ \theta $ gives approximately 20.9°. For the second minimum, $ m = 2 $, $ 1400 \sin \theta = 980 $, solving for $ \theta $ gives approximately 44°.

A Single Slit,1400 Nm Wide,forms a Diffraction Pattern When Illuminated