Physic Labs

Problem 3

Monochromatic light of wavelength λ\lambda is incident normally on a single slit of width bb. A parallel screen is at distance LL, with L≫bL\gg b so Fraunhofer diffraction applies. Intensity is observed at a small angle θ\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 θ1\theta_1 of the first dark minimum. (c) Derive its screen displacement y1y_1 and explain why this is a dark fringe rather than a maximum.
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δ=bsin⁡θ,dϕ=2πλδ\delta=b\sin\theta,\quad d\phi=\frac{2\pi}{\lambda}\delta
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The path difference across the slit is bsin⁡θb\sin\theta, which determines the phase difference. The stated Fraunhofer and small-angle approximations are used.

problems.proof.pitfall. The first single-slit minimum satisfies bsin⁡θ=λb\sin\theta=\lambda, not the two-slit condition bsin⁡θ=λ/2b\sin\theta=\lambda/2.