What is the Width of a Single Slit in Diffraction?

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Homework Statement



Printed problem 11 on single-slit Fraunhofer diffraction: derive I(θ) = I(0)(sin β/β)² with β = (kb/2) sin θ, then find the first minimum for a 0.5 mm slit at λ = 650 nm, the first subsidiary maximum, and the half-maximum angular width.


Homework Equations





The Attempt at a Solution



Part(a)
Diffraction is the bending of light about an aperture. To be able to resolve two distinct images, the minimum angle must satisfy the rayleigh criterion. Any angle smaller will hence affect the quality of images.

(b) Proven by integration.

Handwritten notes giving the single-slit intensity I proportional to sin(β)/β with β = (π/λ) w sinθ, and a sketch of the diffraction pattern with a central maximum and side lobes, marked at -λ/w and λ/w.


(c) For observable diffraction, firstly w must be in the same order of magnitude as λ. Secondly, using w*sinθ = nλ, in order to get the first maxima to be 30o, w must be 2λ etc...

(d) since w*sinθ = nλ, the higher wavelengths of light get spread out more than the lower wavelengths. But the central maxima will be a combination of all colors. At certain angles some wavelengths of light overlap where niλi = njλj, i≠j.