Problem 3

Monochromatic light of wavelength

\lambda

is incident normally on a single slit of width

b

. A parallel screen is at distance

L

, with

L\gg b

so Fraunhofer diffraction applies. Intensity is observed at a small angle

\theta

from the normal through the slit centre; ignore the finite screen size and other edge effects. (a) Divide the aperture into elements and state the condition for cancellation of their amplitudes. (b) Find the angle

\theta_1

of the first dark minimum. (c) Derive its screen displacement

y_1

and explain why this is a dark fringe rather than a maximum.

problems.proof.stepOf

problems.proof.prevStep problems.proof.nextStep

\beta=\pi\Rightarrow b\sin\theta_1=\lambda

problems.proof.analysis

The first minimum occurs at $\beta=\pi$ , meaning the path difference across the slit is one wavelength. The stated Fraunhofer and small-angle approximations are used.