Reflection and Refraction — Question 4

Section 33-7 · Physics pages 5498–5499

Question 4 Parallel layers preserve one transverse ray invariant A ray crosses three plane parallel, stationary, transparent layers. Their indices are n1, n2, and n3 and ray angles to the common normal are theta1, theta2, and theta3. The middle layer has thickness d. Assume all refractions propagate. n1 n2 n3 Schematic; labels, not drawing lengths, specify the data. (a) Eliminate the middle index to relate the entry and final angles. (b) Find the transit time through the middle layer and explain why changing n2 can change time even when the final angle is unchanged.
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Question 4 - Solution Parallel layers preserve one transverse ray invariant n1 n2 n3 (a) Method and calculation n1 sin θ1 = n2 sin θ2 = n3 sin θ3 ≡K. θ3 = sin−1(K/n3) . (b) Interpretation and further calculation ℓ2 = d/ cos θ2, t2 = ℓ2/(c/n2). t2 = n2d c √ 1 −K2/n2 2 . Physical check The end direction depends on the conserved transverse phase gradient, while the internal speed and path length depend on the middle index.

Question 4

Parallel layers preserve one transverse ray invariant

A ray crosses three plane parallel, stationary, transparent layers. Their indices are n1, n2, and n3 and ray angles to the common normal are theta1, theta2, and theta3. The middle layer has thickness d. Assume all refractions propagate.

Diagram shown in the original SVG worksheet.

  1. Eliminate the middle index to relate the entry and final angles.

  2. Find the transit time through the middle layer and explain why changing n2 can change time even when the final angle is unchanged.

Question 4 - Solution

Parallel layers preserve one transverse ray invariant

Diagram shown in the original SVG worksheet.

(a) Method and calculation

n1sin⁡θ1=n2sin⁡θ2=n3sin⁡θ3≡K.n_1\sin\theta_1=n_2\sin\theta_2=n_3\sin\theta_3\equiv K.θ3=sin⁡−1(K/n3).\boxed{\theta_3=\sin^{-1}(K/n_3)}.

(b) Interpretation and further calculation

ℓ2=d/cos⁡θ2,t2=ℓ2/(c/n2).\ell_2=d/\cos\theta_2,\quad t_2=\ell_2/(c/n_2).t2=n2dc1−K2/n22.\boxed{t_2=\frac{n_2d}{c\sqrt{1-K^2/n_2^2}}}.

Physical check

The end direction depends on the conserved transverse phase gradient, while the internal speed and path length depend on the middle index.