Light as a Wave — Question 10

Section 35-1 · Physics pages 5730–5731

Question 10 A tilted plane wave has different apparent spatial periods on two axes In a uniform dielectric of index n, a plane wave has phase ϕ = k0n(x cos θ + y sin θ) −ωt. Let λ0 = 600 nm, n = 1.50, and θ = 30◦. Spatial periods along x and y are measured while the other coordinate and time remain fixed. x y k Schematic; labels, not drawing lengths, specify the data. (a) Find both apparent periods and the actual wavelength along the propagation direction. (b) Write the equation of an equal-phase line and determine its slope.
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Question 10 - Solution A tilted plane wave has different apparent spatial periods on two axes x y k (a) Method and calculation λ∥= λ0/n = 400 nm ; Λx = λ0 n cos θ, Λy = λ0 n sin θ. Λx = 461.880 nm, Λy = 800 nm . (b) Interpretation and further calculation ϕ = ϕ∗=⇒x cos θ + y sin θ = ϕ∗+ ωt k0n . dy dx = −cot θ = − √ 3 ; ∇ϕ = k0n(cos θ, sin θ). Physical check The wave vector is perpendicular to each equal-phase line. A period measured along an oblique coordinate axis exceeds the true wavelength.

Question 10

A tilted plane wave has different apparent spatial periods on two axes

In a uniform dielectric of index n, a plane wave has phase ϕ=k0n(xcos⁡θ+ysin⁡θ)−ωt\phi=k_0n(x\cos\theta+y\sin\theta)-\omega t. Let λ0=600nm\lambda_0=600\,\mathrm{nm}, n=1.50n=1.50, and θ=30∘\theta=30^\circ. Spatial periods along x and y are measured while the other coordinate and time remain fixed.

Diagram shown in the original SVG worksheet.

  1. Find both apparent periods and the actual wavelength along the propagation direction.

  2. Write the equation of an equal-phase line and determine its slope.

Question 10 - Solution

A tilted plane wave has different apparent spatial periods on two axes

Diagram shown in the original SVG worksheet.

(a) Method and calculation

λ∥=λ0/n=400nm;Λx=λ0ncos⁡θ,Λy=λ0nsin⁡θ.\boxed{\lambda_{\parallel}=\lambda_0/n=400\,\mathrm{nm}};\quad\Lambda_x=\frac{\lambda_0}{n\cos\theta},\quad\Lambda_y=\frac{\lambda_0}{n\sin\theta}.Λx=461.880nm,Λy=800nm.\boxed{\Lambda_x=461.880\,\mathrm{nm},\quad\Lambda_y=800\,\mathrm{nm}}.

(b) Interpretation and further calculation

ϕ=ϕ*⇒xcos⁡θ+ysin⁡θ=ϕ*+ωtk0n.\phi=\phi_*\Longrightarrow x\cos\theta+y\sin\theta=\frac{\phi_*+\omega t}{k_0n}.dydx=−cot⁡θ=−3;∇ϕ=k0n(cos⁡θ,sin⁡θ).\boxed{\frac{dy}{dx}=-\cot\theta=-\sqrt3};\quad\nabla\phi=k_0n(\cos\theta,\sin\theta).

Physical check

The wave vector is perpendicular to each equal-phase line. A period measured along an oblique coordinate axis exceeds the true wavelength.