Reflection and Refraction — Question 1

Section 33-7 · Physics pages 5492–5493

Question 1 A wavelength change at a boundary leaves optical frequency unchanged A monochromatic beam travels from index n1=1.20 into index n2=1.60 at 40 degrees to the normal. Its wavelength in the first medium is 600 nm. The media are stationary, transparent, and nondispersive at this frequency. Use c=3.00 times 10 to the eighth m/s. n2 n1 normal reflected Schematic; labels, not drawing lengths, specify the data. (a) Find the refracted and reflected angles. (b) Find the frequency, wavelength in medium 2, and phase-speed ratio.
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Question 1 - Solution A wavelength change at a boundary leaves optical frequency unchanged n2 n1 normal reflected (a) Method and calculation n1 sin θ1 = n2 sin θ2 ⇒θ2 = sin−1(0.75 sin 40◦) = 28.822◦. θr = 40◦. (b) Interpretation and further calculation f = c n1λ1 = 4.16667 × 1014 Hz . λ2 = λ1n1/n2 = 450 nm, v2/v1 = n1/n2 = 0.750 . Physical check A stationary interface preserves temporal frequency; the smaller phase speed produces a shorter wavelength.

Question 1

A wavelength change at a boundary leaves optical frequency unchanged

A monochromatic beam travels from index n1=1.20 into index n2=1.60 at 40 degrees to the normal. Its wavelength in the first medium is 600 nm. The media are stationary, transparent, and nondispersive at this frequency. Use c=3.00 times 10 to the eighth m/s.

Diagram shown in the original SVG worksheet.

  1. Find the refracted and reflected angles.

  2. Find the frequency, wavelength in medium 2, and phase-speed ratio.

Question 1 - Solution

A wavelength change at a boundary leaves optical frequency unchanged

Diagram shown in the original SVG worksheet.

(a) Method and calculation

n1sin⁡θ1=n2sin⁡θ2⇒θ2=sin⁡−1(0.75sin⁡40∘)=28.822∘.n_1\sin\theta_1=n_2\sin\theta_2\Rightarrow\boxed{\theta_2=\sin^{-1}(0.75\sin40^\circ)=28.822^\circ}.θr=40∘.\boxed{\theta_r=40^\circ}.

(b) Interpretation and further calculation

f=cn1λ1=4.16667×1014Hz.f=\frac{c}{n_1\lambda_1}=\boxed{4.16667\times10^{14}\,\mathrm{Hz}}.λ2=λ1n1/n2=450nm,v2/v1=n1/n2=0.750.\boxed{\lambda_2=\lambda_1n_1/n_2=450\,\mathrm{nm},\quad v_2/v_1=n_1/n_2=0.750}.

Physical check

A stationary interface preserves temporal frequency; the smaller phase speed produces a shorter wavelength.