// Numbas version: exam_results_page_options {"name": "PA2240 Unit 3 Three Core Exercises", "metadata": {"description": "", "licence": "None specified"}, "duration": 0, "percentPass": "60", "showQuestionGroupNames": false, "shuffleQuestionGroups": false, "showstudentname": true, "question_groups": [{"name": "Group", "pickingStrategy": "all-ordered", "pickQuestions": 1, "questionNames": ["", "", "", ""], "variable_overrides": [[], [], [], []], "questions": [{"name": "2240_U3_Seminar_Q1", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Martin Barstow", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/5191/"}], "tags": [], "metadata": {"description": "", "licence": "None specified"}, "statement": "

A plane wave in free space is normally incident on a dielectric.

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Show that at the boundary, the components of the magnetic intensity along the surface in the 2 media are continuous.

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(hint: The result is an immediate consequence of the Maxwell equation involving the curl of H. Follow similar working to showing that the E field is continuous)

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Provide the missing part of the equation in the answer.

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$\\rm \\nabla \\times H =$ [[0]]

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Notes for entering symbols in the test:

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subscripts - use _, superscripts - use ^ e.g., E_0T or e^k

\n

multiplication - use *, division - use /

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for a vectors either -

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use m:v: followed by the relevant letter, e.g. m:v:H to get a bold symbol - or -

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use vec: followed by the relevant letter, e.g., vec:E to get ~ over an italic letter

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for greek letters - use their names, e.g., theta, omega

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if using a greek letter in a derivative, leave a space, e.g., d theta

\n

for the operator Del, just type Del

\n

for cross product just type * or leave a space

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Let's consider a rectangular circuit $\\Gamma$ with two sides orthogonal to the surface and two sides parallel to it; by using Stokes's theorem, we get:

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$\\rm  \\oint_\\Gamma $[[0]] $ . \\it d\\bf{l} = \\int_{\\cal S}$ [[1]] $\\cdot d {\\bf s}=\\int_{\\cal S} \\frac{\\partial {\\bf D}}{\\partial t}\\cdot d{\\bf s} $

\n

Again, fill in the missing parts of the equation.

\n

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If we shrink the circuit so that the two sides perpendicular to the surface become infinitesimal we get:

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$({H}_{\\parallel 1} - {H}_{\\parallel 2}) \\Delta l=\\int_{\\cal S} \\nabla \\times {\\bf H}\\cdot d {\\bf s}=$[[0]] 

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where ${H}_{\\parallel 1}$ and ${H}_{\\parallel 2}$ are the components of the H-field parallel inside and outside of the dielectric.

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A laser beam, having a power of 100 MW and a diameter of 1 mm, passes through a

\n

glass window of refractive index 1.59.

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Find the amplitude of the electric and magnetic fields of the incident laser beam in the air.

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Fill the numerical answers in the gaps

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The amplitude of the incident Poynting vector in air is:

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(Note: to enter scientific notation such as $4.6 \\times 10^{-5}$, write 4.5e-5)

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$\\left|\\langle {\\bf S}_{air} \\rangle\\right| =\\frac{P_t}{Area_{laser}}=\\frac{10^8 W}
{\\pi 10^{-6}/4 m^2}\\approx $ [[0]]$\\; W/m^2=\\frac{E_{air}^2}{2Z_0}$

and therefore:

$E_{0_I}[air]=\\sqrt{2\\;Z_0\\;\\langle {\\rm \\bf S}_{air}\\rangle }= $[[1]]$\\:V/m$

and the magnetic field will be

$B_{0_I}[air]=E_{0_I}[air]/c= $[[2]]$T$

\n

\n

Notes for entering symbols in the test:

\n

subscripts - use _, superscripts - use ^ e.g., E_0T or e^k

\n

multiplication - use *, division - use /

\n

for a vectors either -

\n

use m:v: followed by the relevant letter, e.g. m:v:H to get a bold symbol - or -

\n

use vec: followed by the relevant letter, e.g., vec:E to get ~ over an italic letter

\n

for greek letters - use their names, e.g., theta, omega

\n

if using a greek letter in a derivative, leave a space, e.g., d theta

\n

for the operator Del, just type Del

\n

for cross product just type * or leave a space

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Find the amplitude of the electric and magnetic fields of the reflected laser beam in the air:

\n

Fill the numerical answers in the gaps.

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At normal incidence for an air-glass interface:

\n

$\\rm \\frac{E_{0_R}[air]}{E_{0_I}[air]}=\\frac{1-n}{1+n}=$[[0]] 

\n

and therefore:

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$\\rm E_{0_R}[air]=-0.2278 \\:E_{0_I}[air]=$[[1]] V/m

\n

whereas the amplitude of the magnetic field is:

\n

$\\rm B_{0_R}[air]=\\frac{E_{0_R}[air]}{c}=$[[2]] T 

\n

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Find the amplitude of the electric and magnetic fields  of the transmitted laser beam in the glass. 

\n

Fill the numerical answers in the gaps.

\n

At normal incidence for an air-glass interface:

$\\rm \\frac{E_{0_T}[glass]}{E_{0_I}[air]}=\\frac{2}{1+n}=$[[0]] 

and therefore

\n

$\\rm E_{0_T}[glass]=$[[1]] V/m

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whereas the amplitude of the magnetic field is:

$\\rm B_{0_T}[glass]=\\frac{E_{0_T}[glass]}{v_{glass}}=\\frac{E_{0_T}[glass]}{c}n_{glass}=$ [[2]] T

\n

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Compute the reflectance and the transmittance of the beam at the glass interface.

\n

The reflectance is:

$\\rm R=\\left(\\frac{E_{0_R}[air]}{E_{0_I}[air]}\\right)^2=(-0.2278)^2=$[[0]] % 

\n

and the transmittance is:

\n

$\\rm T=1-R=$[[1]] %.

\n

\n

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An electromagnetic wave travelling in a vacuum is normally incident on a slab of glass of refractive index. $n$, and of relative permeability, $\\mu = 1$.

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Write down expressions for the incident, transmitted and reflected E for angular frequency $\\omega $. What is the wavenumber $k$ in each medium in terms of $\\omega $ and $n$.

\n

\n

Incident - $E_I(r,t) = E_{0I}e^{i(k_I.r-\\omega _Rt)}$, $k_I=$[[0]]

\n

Reflected - $E_R(r,t) = E_{0R}$ [[1]] , $k_R=n_1{\\omega _I/c}$

\n

Transmitted - $E_T(r,t) = E_{0T}e^{i(k_T.r-\\omega _Rt)}$, $k_T=$[[2]]

\n

Notes for entering symbols in the test:

\n

subscripts - use _, superscripts - use ^ e.g., E_0T or e^k

\n

multiplication - use *, division - use /

\n

for a vectors either -

\n

use m:v: followed by the relevant letter, e.g. m:v:H to get a bold symbol - or -

\n

use vec: followed by the relevant letter, e.g., vec:E to get ~ over an italic letter

\n

for greek letters - use their names, e.g., theta, omega

\n

if using a greek letter in a derivative, leave a space, e.g., d theta

\n

for the operator Del, just type Del

\n

for cross product just type * or leave a space

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Using the two boundary conditions for tangential fields, give the ratio of the reflected electric field, $E_{0R}$, to the incident electric field, $E_{0I}$ in terms of the impedances of the vacuum and glass, $Z_0$ and $Z_g$, respectively. 

\n

x-component of E is continuous:

\n

$(\\tilde E_{0I} + \\tilde E_{0R})cos\\theta _I = \\tilde E_{0T}$[[0]]

\n

At normal incidence $cos\\theta_I=cos\\theta_T=1$

\n

Therefore:

\n

$(\\tilde E_{0I} + \\tilde E_{0R}) = \\tilde E_{0T}$ 

\n

y-component of H is continuous:

\n

$\\tilde H_{0I}-$[[1]] $=\\tilde H_{0T}$

\n

Impedance $|Z| = $[[2]]

\n

$|\\tilde E_{0R}|/|\\tilde E_{0I}|=$[[3]]$/(Z_g+Z_0)$

\n

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A uniform plane wave travelling in a medium with refractive index $n_1$, is incident on a second medium of refractive index $n_2$ at angle $\\theta_I$.

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Show that

$r =\\left( {\\frac{{\\tilde{E}_{0_R} }}{{\\tilde{E}_{0_I} }}} \\right) = \\frac{{Z_2 \\cos \\theta_I - Z_1 \\cos \\theta _T }}{{Z_2 \\cos \\theta _I + Z_1 \\cos \\theta _T }}$

\n

where $\\tilde{E}_{0_R}$ and $\\tilde{E}_{0_I}$ are the complex magnitudes of the reflected and incident electric fields, $Z_1$ and $Z_2$ are the characteristic impedances of medium 1 and 2, and $\\theta_T$ is the angle of the transmitted wave to the normal.

\n

\n

By imposing the continuity of the tangential components of the E and H fields we get:

\n

(enter the missing parts of the equations)

\n


$\\tilde{E}_{0_I}+\\tilde{E}_{0_R} = $ [[0]]

\n


$ \\tilde{H}_{0_I}\\cos\\theta_I- $ [[1]]$ =  \\tilde{H}_{0_T} \\cos\\theta_T $

\n

By using the relation between the ${\\bf E}$ and ${\\bf H}$ fields via the impedances of the two media we get:

$\\frac{\\tilde{E}_{0_I}}{Z_1}\\cos\\theta_I- $ [[2]]$\\ cos\\theta_I= \\frac{\\tilde{E}_{0_T}}{Z_2} $ [[3]]

and by solving the system of two equations in two unknowns we get:

$ r =\\left( {\\frac{{\\tilde{E}_{0_R} }}{{\\tilde{E}_{0_I} }}} \\right) = \\frac{{Z_2 \\cos \\theta _I - Z_1 \\cos \\theta _T }}{{Z_2 \\cos \\theta _I + Z_1 \\cos \\theta _T }} $

\n

\n

Notes for entering symbols in the test:

\n

subscripts - use _, superscripts - use ^ e.g., E_0T or e^k

\n

multiplication - use *, division - use /

\n

for a vectors either -

\n

use m:v: followed by the relevant letter, e.g. m:v:H to get a bold symbol - or -

\n

use vec: followed by the relevant letter, e.g., vec:E to get ~ over an italic letter

\n

for greek letters - use their names, e.g., theta, omega

\n

if using a greek letter in a derivative, leave a space, e.g., d theta

\n

for the operator Del, just type Del

\n

for cross product just type * or leave a space

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Show that at normal incidence for dielectrics the previous formula reduces to:

\n

$\\frac{{\\tilde{E}_{0_R} }}{{\\tilde{E}_{0_I} }} = \\frac{{n_1-n_2}}{{n_1+n_2 }}$

\n

This comes immediately from the previous result with $\\theta _I = \\theta _R = \\theta _T = $[[0]], and $Z_1=Z_0/n_1$ plus
$Z_2=Z_0/$[[1]], which are valid for dielectrics.

\n

\n

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