A slab of refractive index 1.5 and thickness t is surrounded by air on both sides. Light of free-space wavelength 600 nm is incident normally. Find the minimum nonzero t giving destructive interference in reflected light, in nm.
Reflection at the first air-to-slab surface introduces a π phase shift, while reflection at the slab-to-air surface does not. The propagation phase difference is $4\pi nt/\lambda$. Destructive reflected interference therefore occurs for $2nt=m\lambda$. Excluding zero thickness, take m = 1: $t=600/[2(1.5)]=200$ nm. The quarter-wave thickness instead gives constructive reflection for this low–high–low arrangement.
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