A single thin slit forms a diffraction pattern,with the first minimum at an angle of 40.0° from central maximum.If monochromatic light of 530.0-nm wavelength is used for the illumination,what is the width of the slit?
A)825 nm
B)791 nm
C)757 nm
D)723 nm
E)689 nm

Question

Quizplus · Accepted Answer

The first minimum in a single-slit diffraction pattern occurs when the path length difference between the rays from the edges of the slit is half a wavelength. In this case, we have:$$d\sin	heta = \frac{\lambda}{2}$$where d is the slit width, θ is the angle of the first minimum, and λ is the wavelength of light.Solving for d, we get:$$d = \frac{\lambda}{2\sin	heta}$$Plugging in the given values, we get:$$d = \frac{530.0	ext{ nm}}{2\sin(40.0^\circ)} = 791	ext{ nm}$$Therefore, the correct answer is B.

A Single Thin Slit Forms a Diffraction Pattern,with the First