Съдържание

Условие A. An interesting prism B. Refraction, but…mostly reflection Решение

RMPh 2017, Теория, пролетна сесия — Задача 1. Reflection and refraction of light

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от RMPh 2017

Romanian Master of Physics 2017

Внимание: Бележка към темата

The archive catalogue attaches T1_sol.pdf to the unrelated October 2017 problem “Two paradoxes”. Its nine original pages instead contain the full author solution to this spring problem, “Reflection and refraction of light”: A1–A3, B1, B2.1–B2.4, B2.5.1 and B2.5.2. The printed key has been matched to this existing whole problem. Its numerical values, notation and source defects are retained literally.

Условие

A. An interesting prism

The main section of a glass prism, situated in air $(n' = 1.00)$ , has the form of a rhomb with $\sphericalangle BAD = \sphericalangle BCD \equiv \theta$ . A thin yellow beam of monochromatic light, propagating towards the prism, parallel with the diagonal AC of the rhomb, is incident on the face AB (Fig. 1). The beam is totally reflected on the faces AD and DC, then emerges through its face BC. For the yellow radiation, the refraction index of the glass is $n = 1.60$ .

Rhomb-shaped prism ABCD (cyan) with vertices A (left), B (top), C (right), D (bottom), dashed diagonals AC and BD, angle θ marked at A. A horizontal ray enters face AB, is reflected at faces AD and DC, and exits through face BC horizontally.

Fig. 1

A1 Derive the mathematical expression for the angle $\theta$ as a function of the refraction index $n$ of the prism, such that the total deviation of the beam that exits the prism to be zero. Under the above condition, calculate the numerical value of $\theta$ in degrees and minutes, if $n = 1.60$ . [2,25 т.]

The prism with $\theta$ determined above and the direction of the incident beam remain fixed, but the nature of the light radiation changes, being formed now of the yellow doublet of the mercury. The two wavelengths have the values 579.1 nm, respectively 577 nm. The refraction indices of the glass for these wavelengths are $n = 1.60$ , respectively $n + \Delta n$ , where $\Delta n = 1.3 \cdot 10^{-4}$ . The light rays that exit the prism enter longitudinally into an astronomical telescope adjusted for infinite distance.

A2 Derive the mathematical expression for „the angular distance” $\varepsilon$ between the two images seen through the telescope (first as a function of $\theta$ and $\Delta n$ , then as a function of $n$ and $\Delta n$ ) and calculate its numerical value. [2 т.]

A3 If the focal distance of the telescope’s objective is $f_{ob} = 0.40\ \mathrm{m}$ , derive the linear distance $y$ between the two images, seen in the focal plane of the objective and calculate its numerical value. [0,5 т.]

B. Refraction, but…mostly reflection

B1. Total reflection in geometrical optics

Total reflection occurs when light travels from a medium with refractive index $n_1$ to another one, with the refractive index $n_2 < n_1$ , at an incidence angle $\theta_1 \geq l$ , where $l$ is the critical value of the incidence angle, called limit angle, beyond which there will be no refracted light. At total reflection, the entire energy of the incident light beam goes to the reflected beam.

B1 Derive the mathematical expression for the limit angle [0,25 т.]

B2. Total reflection in electromagnetic optics

Electromagnetic optics proves that, besides being totally reflected, the incident light beam also penetrates the less refringent medium as an evanescent wave.

The characteristics of the reflected and the refracted light beams depend on the angle of incidence, as well as on the orientation of the electric field of the light wave (called polarization). For simplicity, let us consider that the electric field intensity is perpendicular on the incidence plane, as represented in Fig. 2. The indices $i$ , $r$ , and $t$ refer to the incident, reflected and transmitted properties of light waves while $\vec{k}$ is the wave vector, giving the light propagation orientation. Moreover, $\hat{x}$ , $\hat{y}$ and $\hat{z}$ are the unit vectors of the chosen Cartesian reference frame.

Plane interface (dashed horizontal line) between medium n1 (above, white) and medium n2 (below, cyan), with dashed normal. Incident wave vector k_i with field E_i and reflected wave vector k_r with field E_r make angles θ1 with the normal; the transmitted wave vector k_t with field E_t makes angle θ2 with the normal. The fields are marked as crossed circles (perpendicular to the plane). Unit vectors x̂ (into the page), ŷ (right along the interface) and ẑ (down) are shown on the right.

Fig. 2

Physical note: The perturbation produced by a plane, monochromatic wave in a point in space at a certain moment of time can be written as $\vec{E}(\vec{r},t) = \vec{E}_0 \cos(\omega t - \vec{k}\cdot\vec{r})$ , or, to simplify calculations, in the complex form $\vec{e}(\vec{r},t) = \vec{E}_0 e^{i(\omega t - \vec{k}\cdot\vec{r})}$ , where $i = \sqrt{-1}$ , and then taking only the real part of the result.

Mathematical note: For the complex number $z = a \pm ib$ , $a$ is the real part and $b$ is the imaginary part. It can be written as

z = \underbrace{\sqrt{a^2+b^2}}_{=|z|}\left(\underbrace{\frac{a}{\sqrt{a^2+b^2}}}_{=\cos\varphi} \pm i\underbrace{\frac{b}{\sqrt{a^2+b^2}}}_{=\sin\varphi}\right) = |z|e^{\pm i\varphi},

where $|z|$ is the modulus of the complex number $z$ and $\tan\varphi = b/a$ .

B2.1. Evanescent wave

B2.1 Knowing that the incident wave is a plane and monochromatic one, characterized by the equation $\vec{e}_i(\vec{r},t) = \vec{E}_{0i} e^{i(\omega t - \vec{k}_i\cdot\vec{r})}$ , prove that the mathematical expression for the evanescent wave is $\vec{e}_t(\vec{r},t) = \vec{E}_{0t} e^{-\alpha z} e^{i\varphi}$ and derive the exact expression for the attenuation coefficient $\alpha$ as a function of the incidence angle $\theta_1$ , the limit angle $l$ , and the wavelength $\lambda$ of the incident wave. Also, derive the exact expression for the phase $\varphi$ of the evanescent wave. [1,5 т.]

B2.2. Penetration depth

B2.2 Derive the mathematical expression of the distance $\Delta z$ from the interface at which the amplitude of the evanescent wave is $e$ times smaller than at the interface, as a function of the incident wavelength $\lambda$ and calculate its numerical value. The first medium is glass $(n_1 = 1.6)$ , the second is air $(n_2 = 1.0)$ , and the incidence angle of light is $\theta_1 = 40^{\circ}$ . [0,5 т.]

B2.3. The phase speed of the evanescent wave

B2.3 Derive the mathematical expression for the ratio $\dfrac{v_e}{v_1}$ , where $v_e$ is the phase speed of the evanescent wave and $v_1$ - the phase speed of the incident wave, and compute its numerical value for the case of the incidence angle of light of $\theta_1 = 40^{\circ}$ . [0,75 т.]

B2.4. The energy transferred from the incident wave to the totally reflected wave

For any value of the incidence angle, the relationship between the amplitude of the field of the reflected wave and that of the incident wave was derived by the French physicist Augustin Fresnel (1788 – 1829):

E_{0r} = \frac{n_1\cos\theta_1 - n_2\cos\theta_2}{n_1\cos\theta_1 + n_2\cos\theta_2}E_{0i}.

Physical note: If the perturbation produced by a wave in a point in space at a given moment is expressed using complex numbers, then the wave intensity has the mathematical expression $I = \frac{1}{2}\varepsilon_0 c E_0^{*} E_0 = \frac{1}{2}\varepsilon_0 c |E_0|^2$ , where $E_0^{*} = a - ib$ is the complex conjugate of the complex number $E_0 = a + ib$ . Here $\varepsilon_0$ is the vacuum permittivity and $c$ is the speed of light in vacuum.

B2.4 Prove that the totally reflected wave has the same intensity as the incident wave. [0,5 т.]

B2.5. The Goos – Hänchen effect

When an incident wave beam with a finite cross section undergoes total reflection at an interface between two media, the totally reflected wave beam is laterally displaced, on a distance $D$ (see Fig. 3), that was measured for the first time by Goos and Hänchen in 1947. In Fig. 3, the displacement along the surface is $s$ , and the Goos – Hänchen shift is the lateral shift $D$ indicated in the diagram. This is the Goos – Hänchen effect. The explanation of this lateral shift is based on the appearance of the evanescent wave at the interface and its propagation parallel to the interface.

Goos–Hänchen effect: two parallel incident rays in medium n1 hit the interface with medium n2 (cyan) at angle θ1 to the dashed normal; each reflected ray emerges displaced along the interface by s (small dashed arcs below the interface), and the lateral shift D between the geometric reflected ray (dashed) and the actual reflected ray is marked. Unit vectors x̂ (into the page), ŷ (right) and ẑ (down) are shown on the right.

Fig. 3

B2.5.1. The lateral shift

B2.5.1 Derive the mathematical expression for the Goos – Hänchen lateral shift $D$ , admitting that the phase difference between the totally reflected wave and the incident one is zero at the interface. Consequently, compute the numerical value of the displacement $s$ along the interface as a function of the wavelength $\lambda$ of the incident light, if the first medium is glass $(n_1 = 1.6)$ , the second is air $(n_2 = 1.0)$ , and the incidence angle of light is $\theta_1 = 40^{\circ}$ . [1 т.]

B2.5.2. Time needed for the total reflection

An alternative explanation of the Goos – Hänchen shift can be given in terms of the time delay associated with the scattering of a radiation pulse at the interface. The incident radiation pulse is not scattered instantaneously by the surface, but reemerges into medium 1 after a time delay $\tau$ , during which the pulse propagates parallel to the surface and is displaced by the distance $s$ .

B2.5.2. Derive the mathematical expression for the time delay $\tau$ and calculate its value if the first medium is glass $(n_1 = 1.6)$ , the second is air $(n_2 = 1.0)$ , the incidence angle of light is $\theta_1 = 40^{\circ}$ , and the monochromatic radiation has the wavelength $\lambda = 579.1\ \mathrm{nm}$ . The light speed in vacuum is $c = 3.0 \cdot 10^{8}\ \mathrm{m/s}$ . [0,75 т.]

proposed by

Prof. Florea ULIU, PhD

Department of Physics, University of Craiova, ROMANIA

Assoc. Prof. Sebastian POPESCU, PhD

Faculty of Physics, Alexandru Ioan Cuza University of Iași, ROMANIA

Решение

A. An interesting prism

Native rhombus ABCD with incident horizontal ray, two internal reflections and emergent horizontal ray; complete corner labels and theta.

A1

The total deviation is zero only if $I'R\parallel SI$ . This means that $i'=i$ and $r'=r$ (symmetrical path inside of the rhombic prism). Therefore $JJ'\parallel AC\parallel SI\parallel I'R$ . (0.25 p)

From the figure below, $m(\angle J'JD)=m(\angle JJ'D)=\theta/2$ and $m(\angle AJI)=m(\angle CJ'I')=\theta/2$ .

In the triangle $AIJ$ : $\theta+\pi/2+r+\theta/2=\pi$ , so that $r=\pi/2-3\theta/2$ . (0.25 p)

Native complete ray geometry S,I,A,B,C,D,J,J′,I′,R, theta halves, i and i′, r and r′, alpha and beta, normals and angular arcs.

On the other hand, $i=\pi/2-\theta/2$ . (0.25 p)

The refraction law (Snell–Descartes) is $\sin(\pi/2-\theta/2)=n\sin(\pi/2-3\theta/2)$ , or $\cos(\theta/2)=n\cos(3\theta/2)$ . (0.50 p)

Since

\cos(3\theta/2)=\cos(\theta+\theta/2)=\cos\theta\cos(\theta/2)-\sin\theta\sin(\theta/2)

=[2\cos^2(\theta/2)-1]\cos(\theta/2)-2\sin^2(\theta/2)\cos(\theta/2)=\ldots=4\cos^3(\theta/2)-3\cos(\theta/2),

the refraction law becomes $\cos(\theta/2)\{1-n[4\cos^2(\theta/2)-3]\}=0$ . The solution $\theta=\pi$ (i.e. $180^\circ$ ) cannot be accepted as unphysical, so the physical solution is

\cos(\theta/2)=\sqrt{\frac{3n+1}{4n}}=0.952.\qquad\text{(0.75 p)}

i.e., $\theta=35^\circ40'$ . (0.25 p)

A2

We have a fixed angle of incidence $(i=\text{constant})$ and two angles of refraction $r$ and $r-dr$ , corresponding to $n$ and $n+dn$ . From $\sin i=n\sin r$ we can write $0=dn\sin r+n\cos r\,dr$ .

On the other hand, from $\sin i'=n\sin r'$ , we can write $\cos i'\,di'=dn\sin r'+n\cos r'\,dr'$ .

Because $dn$ (or $\Delta n$ , as in the text of the question) is a very small (infinitesimal) quantity, we can approximate $i'=i$ and $r'=r$ , but with $di'\ne di\ (=0)$ and $dr'\ne dr\ (\ne0)$ . (0.50 p)

We will demonstrate later (see Note below) that $dr'=-dr$ .

Now, let us determine the angle between the two emergent rays of light, namely $\varepsilon\equiv di'$ , obtaining

\varepsilon=\frac{\sin r}{\cos i}\,dn+n\frac{\cos r}{\cos i}(-dr),\qquad dr=-\tan r\frac{dn}{n}.

The final result is

\varepsilon=2\frac{\sin r}{\cos i}\,dn=2\left[\frac{\cos(3\theta/2)}{\sin(\theta/2)}\right]dn.\qquad\text{(0.50 p)}

We know the mathematical expression of $\cos(\theta/2)$ and, after a little algebra, $\varepsilon$ can be expressed in another form, namely

\varepsilon=2\frac{dn}{n}\sqrt{\frac{3n+1}{n-1}}.\qquad\text{(0.50 p)}

Numerically: $\varepsilon=5.1\cdot10^{-4}\ \mathrm{rad}$ (or $1.7'$ ). (0.25 p)

Note: In the triangle $JDJ'$ , we write that the sum of inner angles is $\pi$ (radians), namely $(\pi/2-\alpha)+(\pi-\theta)+(\pi/2-\beta)=\pi$ , i.e. $\alpha+\beta=\pi-\theta=\text{constant}$ (for a given rhombic prism, with fixed $\theta$ ). Therefore $d\alpha=-d\beta$ , where $\alpha=\theta+r$ (see the triangle $AIJ$ ) and $\beta=\theta+r'$ (see the triangle $CI'J'$ ). Since $\theta$ is fixed (our situation) $d\alpha=dr$ and $d\beta=dr'$ so that $dr=-dr'$ . (0.25 p)

A3

y=f_{ob}\tan\varepsilon\approx\varepsilon f_{ob}=2\left(\frac{dn}{n}\right)f_{ob}\sqrt{\frac{3n+1}{n-1}}.

Numerically: $y=0.20\ \mathrm{mm}$ . (0.50 p)

B. Refraction, but…mostly reflection

B1. Total reflection in geometrical optics

When the incidence angle $\theta_1$ reaches the limit angle $l$ , the refraction angle will be $\theta_2=90^\circ$ . In this case, the second Snell’s refraction law gives $n_1\sin l=n_2$ , so

l=\arcsin\frac{n_2}{n_1}.

B2. Total reflection in electromagnetic optics

Electromagnetic optics proves that, besides being totally reflected, the incident light beam also penetrates the less refringent medium as an evanescent wave.

Mathematical note: For the complex number $z = a \pm ib$ , $a$ is the real part and $b$ is the imaginary part. It can be written as

z = \underbrace{\sqrt{a^2+b^2}}_{=|z|}\left(\underbrace{\frac{a}{\sqrt{a^2+b^2}}}_{=\cos\varphi} \pm i\underbrace{\frac{b}{\sqrt{a^2+b^2}}}_{=\sin\varphi}\right) = |z|e^{\pm i\varphi},

where $|z|$ is the modulus of the complex number $z$ and $\tan\varphi = b/a$ .

Incident, reflected and transmitted wave vectors and electric fields, n1/n2 interface, theta1/theta2, dashed normal and all three coordinate unit vectors.

B2.1. Evanescent wave

For the transmitted wave

\vec e_t(\vec r,t)=\vec E_{0t}e^{i(\omega t-\vec k_t\cdot\vec r)},

where

\vec k_t\cdot\vec r=(\hat y k_t\sin\theta_2+\hat z k_t\cos\theta_2)\cdot(\hat y y+\hat z z)=k_t(y\sin\theta_2+z\cos\theta_2).

Since

n_1\sin\theta_1=n_2\sin\theta_2,

\cos\theta_2=\pm\sqrt{1-\sin^2\theta_2}=\pm\sqrt{1-\frac{n_1^2}{n_2^2}\sin^2\theta_1}=\pm i\sqrt{\frac{n_1^2}{n_2^2}\sin^2\theta_1-1},

because $\theta_1>l$ , under these conditions the electric field of the transmitted wave can be written as

\vec e_t(\vec r,t)=\vec E_{0t}e^{\pm k_tz\sqrt{(n_1^2/n_2^2)\sin^2\theta_1-1}}e^{i(\omega t-k_ty(n_1/n_2)\sin\theta_1)}.

The $+$ sign in the first exponential has no physical significance because there is no wave at appreciable distances from the interface. In conclusion, the electric field of the transmitted (evanescent) wave has the form

\vec e_t(\vec r,t)=\vec E_{0t}e^{-\alpha z}e^{i\varphi},

where $\alpha=k_t\sqrt{(n_1^2/n_2^2)\sin^2\theta_1-1}$ is the attenuation coefficient of the evanescent wave and $\varphi=\omega t-k_ty(n_1/n_2)\sin\theta_1$ is the wave’s phase. This result shows that the wave travels along the interface (along $y$ direction) and that it is attenuated in the $z$ direction (perpendicular on the interface).

Because

k_t=\frac{\omega}{v_2}=\frac{\omega}{v_1}\frac{v_1}{v_2}=k_i\frac{c/n_1}{c/n_2}=k_i\frac{n_2}{n_1},

then

\alpha=k_i\sqrt{\sin^2\theta_1-\frac{n_2^2}{n_1^2}}=\frac{2\pi}{\lambda}\sqrt{\sin^2\theta_1-\frac{n_2^2}{n_1^2}}=\frac{2\pi}{\lambda}\sqrt{\sin^2\theta_1-\sin^2 l}.

B2.2. Penetration depth

\Delta z=\frac1\alpha=1.1\lambda.

B2.3. The phase speed of the evanescent wave

Considering the phase of the evanescent wave, its phase speed is

v_e=\frac{\omega}{k_t(n_1/n_2)\sin\theta_1}=\frac{\omega}{k_i\sin\theta_1}=\frac{v_1}{\sin\theta_1},

so the requested ratio is

\frac{v_e}{v_1}=\frac1{\sin\theta_1}=1.6.

B2.4. The energy transferred from the incident wave to the totally reflected wave

Since

n_2\cos\theta_2=-n_2\sqrt{1-\sin^2\theta_2}=-n_2\sqrt{1-\frac{n_1^2}{n_2^2}\sin^2\theta_1}=-i n_2\sqrt{\frac{n_1^2}{n_2^2}\sin^2\theta_1-1}=-i\frac{n_1\alpha}{k_i},

then

E_{0r}=\frac{n_1\cos\theta_1+i n_1\alpha/k_i}{n_1\cos\theta_1-i n_1\alpha/k_i}E_{0i}=\frac{\cos\theta_1+i\alpha/k_i}{\cos\theta_1-i\alpha/k_i}E_{0i}.

For any complex number of the same form we can write

\frac{a+ib}{a-ib}=\frac{\sqrt{a^2+b^2}\,e^{i\varphi_0}}{\sqrt{a^2+b^2}\,e^{-i\varphi_0}}=e^{2i\varphi_0},\qquad \tan\varphi_0=\frac ba.

In conclusion $E_{0r}=e^{2i\varphi_0}E_{0i}$ , where

\tan\varphi_0=\frac{\alpha}{k_i\cos\theta_1}=\frac1{\cos\theta_1}\sqrt{\sin^2\theta_1-\sin^2 l}.

Under these conditions, the intensity of the totally reflected wave will be

I_r=\frac12\varepsilon_0c|E_{0r}|^2=\frac12\varepsilon_0c|E_{0i}|^2=I_i.

B2.5. The Goos–Hänchen effect

Goos–Hänchen geometry with two complete incident/reflected rays, dashed geometric rays, lateral D and surface s displacements, theta1, n1/n2, and x/y/z unit vectors.

B2.5.1. The lateral shift

From Fig. 3, it follows that $D=s\cos\theta_1$ .

If in the incidence point for the incident wave, on the interface, the phase of its electric field is $\Phi_i$ , then, at the starting point, for the totally reflected wave its phase is

\Phi_r=\Phi_i-k_{iy}s+2\varphi_0.

With $\Phi_r=\Phi_i$ , and knowing that $k_{iy}=k_i\sin\theta_1$ , we obtain

s=\frac{2\varphi}{k_i\sin\theta_1}=\frac{\lambda}{\pi}\frac1{\sin\theta_1}\tan^{-1}\left(\frac{\sqrt{\sin^2\theta_1-\sin^2 l}}{\cos\theta_1}\right).

Finally,

D=\frac{\lambda}{\pi}\frac{\tan^{-1}\left(\sqrt{\sin^2\theta_1-\sin^2 l}/\cos\theta_1\right)}{\tan\theta_1}.

The numerical value of $s$ is $s=5.5\lambda$ .

B2.5.2. Time needed for the total reflection

\tau=\frac{s}{v_{1y}}=\frac{s}{v_1\sin\theta_1}=\frac{sn_1}{c\sin\theta_1}=\frac{\lambda n_1}{\pi c\sin^2\theta_1}\tan^{-1}\left(\frac{\sqrt{\sin^2\theta_1-\sin^2 l}}{\cos\theta_1}\right)=2.6\cdot10^{-14}\ \mathrm{s}.

proposed by

Prof. Florea ULIU, PhD — Department of Physics, University of Craiova, ROMANIA

Assoc. Prof. Sebastian POPESCU, PhD — Faculty of Physics, Alexandru Ioan Cuza University of Iași, ROMANIA

Original blank answer sheet — page 6

The original answer sheet is blank; these are requested answer spaces, not supplied answers.

Task	Mathematical expression	Numerical value
A1	$\theta(n)=$	$\theta(n=1.6)=$
A2	$\varepsilon(\theta,\Delta n)=$ and $\varepsilon(n,\Delta n)=$	$\varepsilon(n,\Delta n)=$
A3	$y(f_{ob},n,\Delta n)=$	$y(f_{ob},n,\Delta n)=$
B1	$l=$

Original blank answer sheet — page 7

Task	Mathematical expression	Numerical value
B2.1	$\alpha(\theta_1,l,\lambda)=$ and $\varphi=$
B2.2	$\Delta z(\lambda)=$	$\Delta z(\lambda)=$
B2.3	$v_e/v_1=$	$v_e/v_1=$

B2.4 has a large empty derivation box which continues onto page 8.

Original blank answer sheet — page 8

Task	Mathematical expression	Numerical value
B2.5.1	$D(\lambda,\theta_1,l)=$	$s(\lambda)=$
B2.5.2	$\tau(\lambda,\theta_1,l,c,n_1)=$	$\tau=$

Оригинал в Архива: RMPh2017Q1ENG.pdf · официални решения: T1_sol.pdf