Light as a Wave — Question 3

Section 35-1 · Physics pages 5716–5717

Question 3 A phase ramp across an aperture steers the emerging wavefront A thin transparent plate in air has thickness t(y) = t0 + ay, index n = 1.50, and slope a = 0.020. A normally incident plane wave acquires transmission phase Φ(y) = k0(n −1)t(y). The outgoing field has phase kxx + kyy −ωt. Neglect reflection and use the local phase-screen model. x y k Schematic; labels, not drawing lengths, specify the data. (a) Determine the outgoing transverse wave-vector component and angle from the positive x axis. (b) Find the transverse distance over which the plate contributes one full phase cycle for λ0 = 500 nm.
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Question 3 - Solution A phase ramp across an aperture steers the emerging wavefront x y k (a) Method and calculation ky = dΦ dy = k0(n −1)a, ky = k0 sin θ. sin θ = 0.010, θ = 0.5730◦. (b) Interpretation and further calculation Φ(y + Λy) −Φ(y) = k0(n −1)aΛy = 2π. Λy = λ0 (n −1)a = 50.0 µm . Physical check The positive thickness gradient produces positive transverse phase gradient under the stated phase convention.

Question 3

A phase ramp across an aperture steers the emerging wavefront

A thin transparent plate in air has thickness t(y)=t0+ayt(y)=t_0+ay, index n=1.50n=1.50, and slope a=0.020a=0.020. A normally incident plane wave acquires transmission phase Φ(y)=k0(n−1)t(y)\Phi(y)=k_0(n-1)t(y). The outgoing field has phase kxx+kyy−ωtk_xx+k_yy-\omega t. Neglect reflection and use the local phase-screen model.

Diagram shown in the original SVG worksheet.

  1. Determine the outgoing transverse wave-vector component and angle from the positive x axis.

  2. Find the transverse distance over which the plate contributes one full phase cycle for λ0=500nm\lambda_0=500\,\mathrm{nm}.

Question 3 - Solution

A phase ramp across an aperture steers the emerging wavefront

Diagram shown in the original SVG worksheet.

(a) Method and calculation

ky=dΦdy=k0(n−1)a,ky=k0sin⁡θ.k_y=\frac{d\Phi}{dy}=k_0(n-1)a,\qquad k_y=k_0\sin\theta.sin⁡θ=0.010,θ=0.5730∘.\boxed{\sin\theta=0.010,\quad\theta=0.5730^\circ}.

(b) Interpretation and further calculation

Φ(y+Λy)−Φ(y)=k0(n−1)aΛy=2π.\Phi(y+\Lambda_y)-\Phi(y)=k_0(n-1)a\Lambda_y=2\pi.Λy=λ0(n−1)a=50.0μm.\boxed{\Lambda_y=\frac{\lambda_0}{(n-1)a}=50.0\,\mu\mathrm m}.

Physical check

The positive thickness gradient produces positive transverse phase gradient under the stated phase convention.