Homework Solutions

255
At the left junction and right junction respectively

1 sin θ_{1} = n sin ψ, n sin ψ = 1 sin θ_{2}

implying $θ_{1} = θ_{2}$ . At the point where the incident ray touches the left wall, drop a perpendicular of length $x$ between entering ray and dotted extrapolation of exiting ray, then with $y$ being the length of the line connecting entry and exit points for the refracted ray, and $d$ being the glass thickness, there are two right triangles of hypotenuse $y$ ;

x = y sin (θ_{1} - ψ), d = y cos ψ

Small angle approximation;

x = y (θ_{1} - ψ), d = y cos ψ \approx y, x = d (θ_{1} - ψ)

insert Snel; $θ_{1} \approx n ψ$ , to get $x = d (θ_{1} - \frac{θ_{1}}{n})$ .

256

sin θ_{2} = n sin ψ_{2}, sin θ_{1} = n sin ψ_{1}

The quadrilateral consisting of the apex of the prism and the two normals to the side surfaces has angles $Φ, 90, 90, 180 - Φ$ , the triangle consisting of the cred refracted ray and the two normals has angles $ψ_{1}, ψ_{2}, 180 - Φ$ and so $ψ_{1} + ψ_{2} = Φ$ . The triangle consisting of dotted rays and the red refracted ray has angles $180 - Ψ, θ_{1} - ψ_{1}, θ_{2} - ψ_{2}$ and therefore

Ψ = θ_{1} + θ_{2} - ψ_{1} - ψ_{2} = θ_{1} + θ_{2} - Φ

When $θ_{1} = θ_{2} = θ$ we have $ψ_{1} = ψ_{2} = ψ$ then

Ψ + Φ = 2 θ, Φ = 2 ψ, sin \frac{Ψ + Φ}{2} = sin θ, sin \frac{Φ}{2} = sin ψ, \frac{sin \frac{Ψ + Φ}{2}}{sin \frac{Φ}{2}} = \frac{sin θ}{sin ψ} = n by Snel

257

258

\frac{R}{2.0 m} = tan θ_{c}, sin θ_{c} = \frac{1}{1.333} = \frac{3}{4}, R = 2.0 m tan θ_{c} = 2.0 m \frac{\frac{3}{4}}{\sqrt{1 - {(\frac{3}{4})}^{2}}} = \frac{6.0 m}{\sqrt{7}}

259

sin θ_{c} = sin (90 - Φ) = \frac{1}{1.52} = cos Φ, Φ = {cos}^{- 1} \frac{1}{1.52} = 48.8 6^{o}

260

sin θ = n_{2} sin ψ, n_{2} sin (90 - ψ) = n_{2} cos ψ = n_{1} sin 90

therefore

sin ψ = \sqrt{1 - {(\frac{n_{1}}{n_{2}})}^{2}}, sin θ = n_{2} sin ψ = n_{2} \sqrt{1 - {(\frac{n_{1}}{n_{2}})}^{2}} = \sqrt{n_{2}^{2} - n_{1}^{2}}

261

θ_{B} = \frac{1.333}{1}, θ_{B} = 53.1 2^{o}

If unpolarized light is incident at this angle, the reﬂected beam is polarized with its electric ﬁeld parallel to the surface of reﬂection. This will depend of $λ$ since in general $n = n (λ)$ as we say in section 20.2.

262
Position “B” will let no light pass. The reﬂected ray has its electric ﬁeld polarized parallel to the “A” arrow.

263

f r a c = \frac{\int_{0}^{θ_{c}} sin θ d θ \int_{0}^{2 π} d φ}{\int_{0}^{π} sin θ d θ \int_{0}^{2 π} d φ} = \frac{1}{2} (- cos θ |_{0}^{θ_{c}}) = \frac{1}{2} (1 - cos θ_{c}) = \frac{1}{2} (1 - \sqrt{1 - n^{- 2}})

264

sin ψ = \frac{h}{R}, n sin ψ = 1 sin 90 = 1, n \frac{h}{R} = 1, h = \frac{R}{n}

265
The electric ﬁeld amplitude of the incident light is $E_{0}$ , $Ī_{0} = 100 \frac{W}{m^{2}} = \frac{1}{2 μ_{0} c} E_{0}^{2}$ . The electric ﬁeld amplitude of the light transmitted through the ﬁrst junction is

E_{T} = \frac{2 \cdot 1}{1 + 3} E_{0} = \frac{2}{4} E_{0}

and transmitted through the second junction

E_{T T} = \frac{2 \cdot 3}{3 + 1} E_{T} = \frac{6}{4} \frac{2}{4} E_{0} = \frac{12}{16} E_{0} = \frac{3}{4} E_{0}

The intensity transmitted through the sheet into the air beyond is

Ī_{T T} = \frac{1}{2 μ_{0} c} E_{T T}^{2} = \frac{3^{2}}{4^{2}} \frac{1}{2 μ_{0} c} E_{0}^{2} = 56.25 \frac{W}{m^{2}}

266

100 = \frac{1}{2 μ_{0} c} E_{i n c}^{2}, I_{t r a n s} = \frac{1}{2 μ_{0} c} (E_{i n c} \frac{2 \cdot 1}{1 + 3} \frac{2 \cdot 3}{1 + 3})^{2} = 0.56 \cdot \frac{1}{2 μ_{0} c} E_{i n c}^{2} = 56 \frac{W}{m^{2}}

267

E_{d i r} = E_{i n c} \frac{2 \cdot 1}{1 + 3} \frac{2 \cdot 3}{1 + 3} = \frac{12}{16} E_{i n c}, E_{s e c} = - E_{i n c} \frac{2 \cdot 1}{1 + 3} \frac{2 \cdot 3}{1 + 3} (\frac{3 - 1}{3 + 1})^{2} = - \frac{3}{16} E_{i n c}

I_{t r a n s} = \frac{1}{2 μ_{0} c} (\frac{12}{16} E_{i n c} - \frac{3}{16} E_{i n c})^{2} = 0.316 \frac{1}{2 μ_{0} c} E_{i n c}^{2} = 31.6 \frac{W}{m^{2}}

Clearly there will be many such beams that need to be taken into account to ﬁnd the total, true transmission through the glass.
The two rays are $18 0^{o}$ out of phase after two reﬂections with no inversions if

2 \cdot 3 D = (m + \frac{1}{2}) λ, D = \frac{λ}{2 \cdot 2 \cdot 3} = 4.17 \times 1 0^{- 8} m

is the minimal glass thickness.

268

1 sin θ = n sin φ, θ \approx n φ, x = D tan φ \approx D φ, x = d tan θ \approx d θ

1 = \frac{D}{d} \frac{φ}{θ} = \frac{D}{d} \frac{1}{n}, d \approx \frac{D}{n}

269

sin θ = \frac{L}{R}, n sin ψ = 1 sin θ, sin ψ = \frac{L}{n R}

but

2 ψ + 180 - θ = 180, ψ = \frac{θ}{2}

therefore

sin θ = \frac{L}{R} = 2 sin \frac{θ}{2} cos \frac{θ}{2} = 2 \frac{L}{n R} \sqrt{1 - {(\frac{L}{n R})}^{2}}

270

(p - 2 (p \cdot n) n) \cdot (p - 2 (p \cdot n) n) = p \cdot p - 4 p \cdot ((p \cdot n) n) + 4 ((p \cdot n) n) \cdot ((p \cdot n) n)

= p^{2} - 4 {(p \cdot n)}^{2} + 4 {(p \cdot n)}^{2} n^{2} = 1 - 4 {(p \cdot n)}^{2} + 4 {(p \cdot n)}^{2} \cdot 1 = 1

since $p^{2} = n^{2}$ since they are ubnit vectors.
For this corner, $n_{1} = - sin θ j + cos θ i$ (right surface) and $n_{2} = + sin φ j + cos φ i$ (left surface). Let the ray hit the right surface ﬁrst. After reﬂection

p = - i \to - i - 2 (- cos θ) (- sin θ j + cos θ i) = cos (2 θ) i - sin (2 θ) j = p^{'}

now let it hit the second surface;

p^{'} \to p^{'} - 2 (p^{'} \cdot n) = cos (2 θ) i - sin (2 θ) j - 2 (cos (2 θ) cos φ - sin (2 θ) sin φ) (sin φ j + cos φ i)

= - cos (2 θ - 2 φ) i - sin (2 θ - 2 φ) j

271
The three transformations

p \to p - 2 (i \cdot p) i, p \to p - 2 (j \cdot p) j, p \to p - 2 (k \cdot p) k

applied to $p = (p_{x}, p_{y}, p_{z})$ in succession perform

(p_{x}, p_{y}, p_{z}) \to (- p_{x}, p_{y}, p_{z}) \to (- p_{x}, - p_{y}, p_{z}) \to (- p_{x}, - p_{y}, - p_{z})

and therefore completely invert the propagation vector.