(a) The irradiance of a single slit of width is , where . By differentiating with…

PHYSICS · Subjective · Fermi University Physics · Optics & Photonics · Interference and diffraction · Interference and diffraction

Optics and Photonics, Ch Fraunhofer diffraction from apertures and gratings

Problem

(a) The irradiance of a single slit of width bb is I(β)=I0(sin⁡ββ)2I(\beta) = I_0\left(\dfrac{\sin\beta}{\beta}\right)^2, where β=πbsin⁡θλ\beta = \dfrac{\pi b\sin\theta}{\lambda}. By differentiating II with respect to β\beta and setting the derivative to zero, show that the secondary maxima (excluding the central maximum at β=0\beta=0) occur where tan⁡β=β\tan\beta=\beta. Taking β1≈4.493\beta_1\approx4.493 rad as the first positive root of this equation, find the fraction of the central peak's irradiance carried by the first secondary maximum, and compare it with the cruder estimate obtained by assuming β≈3π/2\beta\approx3\pi/2. (b) For an NN-slit diffraction grating of slit spacing dd at wavelength λ\lambda, the irradiance is I(δ)=I0(sin⁡(Nδ/2)sin⁡(δ/2))2I(\delta)=I_0\left(\dfrac{\sin(N\delta/2)}{\sin(\delta/2)}\right)^2 with δ=kdsin⁡θ\delta=kd\sin\theta, k=2π/λk=2\pi/\lambda. Starting from the principal maximum at δ=2πm\delta=2\pi m and the zero of the numerator adjacent to it, derive the angular half-width Δθ\Delta\theta (from the maximum to that zero) of the order-mm line, and show that Δθ=λ/(Ndcos⁡θ)\Delta\theta=\lambda/(Nd\cos\theta). (c) A grating with 30003000 lines in total, ruled at spacing d=2.0 μmd=2.0\ \mu\text{m}, is illuminated at normal incidence with λ=500 nm\lambda=500\ \text{nm} and viewed in first order (m=1m=1). Use the result of part (b) to find Δθ\Delta\theta in radians.

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