A beam of light of wavelength $610 \mathrm {~nm}$ passes through a slit that is $1.90 \mu \mathrm { m }$ wide. At what the angle away from the centerline does the first dark fringe occur?

Question

Quizplus · Accepted Answer

The first dark fringe occurs when the path length difference between the two rays from the slit is equal to half a wavelength. The path length difference is equal to $d \sin \theta$, where $d$ is the slit width and $\theta$ is the angle away from the centerline. So, we have:$$d \sin \theta = \frac{\lambda}{2}$$$$\sin \theta = \frac{\lambda}{2d} = \frac{610 \mathrm {~nm}}{2(1.90 \mu \mathrm { m })} = 0.161$$$$\theta = \sin^{-1}(0.161) = 9.35 ^ { \circ }$$

A Beam of Light of Wavelength 610 nm610 \mathrm {~nm}610 nm Passes Through a Slit That Is

A Beam of Light of Wavelength $610 \mathrm {~nm}$ Passes Through a Slit That Is