Съдържание

Условие Part A: Light Propagation Through a Semi-Sphere (1,5 т.)Part B: Light Propagation Through a Cylindrical Rod (3,5 т.)Part C: Light Propagation Through an Optical Fibre (5 т.)Отговори Решение

APhO 2024, Теория — Задача 2. Snell's Law

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Всички материали от APhO 2024

24th Asian Physics Olympiad, Malaysia 2024 — APHO 2024 Theory, English (Official) · 10 т.

General Instructions for Theoretical Examination

Before the Exam

The allocated time for theoretical examination is 5 hours for answering 3 questions that carry 10 points each.
Please ensure that your student code matches the one on your desk. Place your ID / tag on the top right corner of the desk.
Kindly ensure that you have the following items on the desk: (i) A sealed envelope containing the Question Papers and Answer Scripts.
(ii) A labelled envelope.
(iii) Pens/relevant stationery (You need to bring in your own calculator).
If you find that the items are incomplete, kindly raise your hands.
You must not open the sealed envelopes before the start of the examination.

During the Exam

You must use the dedicated answer scripts provided for writing your answers. Write your answers only in boxed answer scripts. If you need more boxed answer scripts, please raise your hand and ask an invigilator. Clearly write your student code and write which question you are attempting on each of your extra requested answer scripts.
There are additional blank draft papers. These will not be marked.
Try to be as concise as possible in your answers: use equations, logical operators, and sketches to illustrate your thoughts whenever possible. Avoid the use of long sentences.
Use an appropriate number of significant figures when stating numbers.
You are not allowed to leave your working place without permission. If you need any assistance (malfunctioning calculator, need to visit a restroom, insufficient Draft Papers, etc), please draw the attention of the invigilator by raising your hand.

At the end of the Exam

At the end of the examination, you must stop writing immediately.
Put all the Answer Scripts into the labelled envelope and seal it, including any blank Answer Scripts. It is your responsibility to ensure all Answer Scripts are inside the envelope and submitted to the invigilator.
Put question papers and draft papers into another envelope. You are not allowed to take any sheets of paper out of the examination hall.
Wait at your table in silence till your envelopes are collected. Once all envelopes are collected, your guide will escort you out of the examination area.

Answer Scripts

You must use the dedicated Answer Scripts provided for writing your answers. Write your answers only in boxed answer scripts. If you need more boxed answer scripts, please raise your hand and ask an invigilator. Clearly write your student code and write which question number you are attempting on each of your extra requested answer scripts.

There are additional blank draft papers. These will not be marked.

Physical Constants

Acceleration due to gravity	$g = 9.81\ \mathrm{m\,s^{-2}}$
Boltzmann constant	$k_B = 1.38 \times 10^{-23}\ \mathrm{J\,K^{-1}}$
Current Mass of the Sun	$M_S = 2.00 \times 10^{30}\ \mathrm{kg}$
Current Radius of the Sun	$R_S = 7.00 \times 10^{8}\ m$
Average Mass of a Galaxy	$M_{galaxy} = 1.5 \times 10^{12}\ M_S$
Magnitude of the electron charge	$e = 1.60 \times 10^{-19}\ \mathrm{C}$
Mass of the electron	$m_e = 9.11 \times 10^{-31}\ \mathrm{kg}$
Mass of the proton	$m_p = 1.67 \times 10^{-27}\ \mathrm{kg}$
Mass of the neutron	$m_n = 1.67 \times 10^{-27}\ \mathrm{kg}$
Permeability of free space	$\mu_0 = 4\pi \times 10^{-7}\ \mathrm{T\,m\,A^{-1}}$
Permittivity of free space	$\varepsilon_0 = 8.85 \times 10^{-12}\ \mathrm{F\,m^{-1}}$
Intrinsic impedance of free space	$Z_0 = 120\pi\ \Omega$
Planck’s constant	$h = 6.63 \times 10^{-34}\ \mathrm{J\,s}$
Speed of light in vacuum	$c = 3.00 \times 10^{8}\ \mathrm{m\,s^{-1}}$
Stefan-Boltzmann constant	$\sigma = 5.67 \times 10^{-8}\ \mathrm{W\,m^{-2}\,K^{-4}}$
Universal Gravitational constant	$G = 6.67 \times 10^{-11}\ \mathrm{N\,m^{2}\,kg^{-2}}$

Условие

Introduction

When light is incident upon a dielectric interface, it will be reflected and refracted, depending on the incident angle of the light and the refractive indices of the dielectric media as shown in Fig. 1. The refraction of light is governed by the Snell’s law,

n_1 \sin \phi_1 = n_2 \sin \phi_2 \qquad (1)

where $n_1$ and $n_2$ are the refractive indices of the lower and upper parts of the boundary, $\phi_1$ and $\phi_2$ are the angles that the light ray makes with the normal of the boundary.

A horizontal boundary with a dashed vertical normal; a ray arrives from the lower medium n1 at angle ø1 to the normal and leaves into the upper medium at a larger angle ø2; the upper-left corner is labelled n2 < n1, the lower-left n1. — Fig. 1: Refraction of light from a dielectric medium with refractive index, $n_1$ , to a second dielectric medium with refractive index, $n_2$ .

Part A: Light Propagation Through a Semi-Sphere (1,5 т.)

Fig. 2 illustrates how light that consists of two rays with different colour, $a$ and $b$ , is incident along the radius of a semi-sphere with refractive index $n_x$ , in air, before being refracted at the bottom at an angle of $\theta_a$ and $\theta_b$ , respectively.

A semi-sphere of refractive index nx with its flat face down; an incoming ray along a radius meets the centre of the flat face at angle θi to the dashed normal and splits below into two refracted rays a (angle θa) and b (angle θb, farther from the normal); the medium outside is labelled n2. — Fig. 2: Light propagation through a semi-sphere.

A.1 Which ray ( $a$ or $b$ ) propagates faster in the semi-sphere? Please provide justification. [0,5 т.]

A.2 When the incident angle, $\theta_i$ , is slowly increased to $45^{\mathrm{o}}$ , ray $b$ no longer exits the bottom of the semi-sphere. When $\theta_i$ is further increased to $50^{\mathrm{o}}$ , the same happens to ray $a$ . What is the difference between the refractive index of ray $a$ and $b$ in the semi-sphere? [1 т.]

Part B: Light Propagation Through a Cylindrical Rod (3,5 т.)

A cylindrical rod has a refractive index of $n_1$ = 1.50. The rod is placed in air, with one end coated with a polymer with refractive index $n_3$ = 1.40, as shown in Fig. 3 below. Light is incident from the polymer into the rod at an angle, $\theta$ . When $\theta$ is changed, there is an instance when light is totally reflected back to the polymer.

A long horizontal rod n1 (blue) in surrounding medium n2, its left end in a green polymer layer n3; a black ray enters from the polymer at angle θ to the dashed axis and zigzags along the rod by total internal reflection, and an orange ray is reflected back into the polymer. — Fig. 3: A cylindrical rod with one end coated with a polymer with refractive index $n_3$ , where $n_2 < n_3 < n_1$ .

B.1 Determine the range of incident angle, $\theta$ , for this condition to happen. [2 т.]

B.2 How does the condition in B.1 change,

(i) if the other, open end of the rod is now coated with a thick layer of oil with refractive index of 1.60? 0.6pt

(ii) if the setup is placed in water with refractive index of 1.33? 0.9pt [1,5 т.]

Part C: Light Propagation Through an Optical Fibre (5 т.)

Optical fibre is formed by surrounding the medium with refractive index, $n_1$ with a lower refractive index medium, $n_2$ , as shown in Fig. 4 below. The medium with refractive index, $n_1$ is known as the fibre core, and the medium with refractive index, $n_2$ is known as the fibre cladding. The refractive index, $n_3$ , is typically the refractive index of air ( $n_3$ = 1).

Left: a 3D sketch of a cylindrical optical fibre with a dashed inner core. Right: a longitudinal cross section with core n1 between cladding layers n2 and outside medium n3; a ray enters the end face at angle θa to the dashed z axis and is reflected at the core–cladding boundary; axes r and z are marked. — Fig. 4: A schematic of an optical fibre and its cross section.

C.1 In order for light to propagate inside the optical fiber, it first needs to be coupled into the optical fiber. The maximum angle of the light incident at the end of the fiber, $\theta_a$ so that the light can be guided inside the fiber without being refracted out is related to the refractive indices of the fiber core and cladding. Consider a ray that propagates along the meridional plane (plane that cuts across the fibre axis), formulate the relationship between the angle, $\theta_a$ and the refractive indices. [2 т.]

C.2 Optical fibers are not always aligned in a straight line but need to be bent to go through different spaces. Given the fiber core diameter of 50 $\mu\mathrm{m}$ , if the minimum bending radius on the optical fiber is 1.0 $\mathrm{cm}$ , how much will $\theta_a$ change? [2,6 т.]

C.3 Given the refractive index of the fiber core and cladding is 1.45 and 1.44, respectively, what are the maximum angle of light incident for case C.1 and C.2 above, when the fiber is placed in air? [0,4 т.]

Отговори

Решение

Part A: Light Propagation Through a Semi-Sphere

A.1 Using Snell’s law, larger angle of refraction corresponds to larger refractive index that the light sees in the semi-sphere ( $n_1$ ). [0.2pt]

This means that the colour with a smaller angle of refraction propagates faster in the semi-sphere, which is colour a. [0.3pt] [0,5 т.]

A.2 The light rays no longer exit the bottom of semi-sphere when the ray is refracted by $90^{\mathrm{o}}$ with respect to the normal. [0.2pt]

Using Snell’s law,

n_x\sin\theta_i = n_2\sin(90^{\circ})

[0.2pt]

Replace “x” with “a” and “b” for the condition of ray a and ray b, respectively.

For ray a,

n_a\sin(50^{\circ}) = 1.0\sin(90^{\circ})

n_a = 1.305

[0.2pt]

For ray b,

n_b\sin(45^{\circ}) = 1.0\sin(90^{\circ})

n_b = 1.414

[0.2pt]

Difference in refractive index, Δn = 0.109 [0.2pt] [1 т.]

Part B: Light Propagation Through a Cylindrical Rod

B.1 Given $n_1$ = 1.5, $n_2$ = 1.0, $n_3$ = 1.4,

A horizontal rod n1 (blue) in surrounding medium n2, its left end in a green polymer layer n3; a black ray enters from the polymer at angle θ to the dashed axis, is refracted at angle θ1, meets the upper surface at point A at angle ϕ1 to the dashed normal and zigzags to the right end B, where it meets the end face at angle θ2; an orange ray travels back and is reflected into the polymer. — Rod with rays for conditions #1 and #2

Identify 2 conditions that need to be obeyed: [0.2pt]

#1: Light incident at point A should be totally reflected.

n_1\sin\phi_1 > n_2\sin(90^{\circ})

\phi_1 > \sin^{-1}\frac{1.0}{1.5} = 41.81^{\circ}

[0.3pt]

The condition for angle $\theta_1 < 90^{\mathrm{o}} - 41.81^{\mathrm{o}} = 48.19^{\mathrm{o}}$ [0.2pt]

Using Snell’s law,

\theta < \sin^{-1}\frac{1.5}{1.4}\sin(48.19^{\circ}) = 53^{\circ}

[0.2pt]

#2: Light incident at point B should be totally reflected.

n_1\sin\theta_2 > n_2\sin(90^{\circ})

\theta_2 > \sin^{-1}\frac{1.0}{1.5} = 41.81^{\circ}

[0.3pt]

The condition for angle $q_1 > 41.81^{\mathrm{o}}$ [0.2pt]

Using Snell's Law

\theta > \sin^{-1}\frac{1.5}{1.4}\sin(41.81^{\circ}) = 45.58^{\circ}

[0.2pt]

Therefore, the incident angle, $q$ , where light is totally reflected back to the polymer is $45.58^{\mathrm{o}} < q < 53^{\mathrm{o}}$ . [0.4pt] [2 т.]

B.2 (i) When right end of rod is coated with oil with $n_4$ = 1.6, condition #2 changes. [0.2pt]

Light incident on point B will always be refracted. [0.2pt]

Light CANNOT be totally reflected back to the polymer. [0.2pt] [0,6 т.]

B.2 (ii) When the setup is placed in water, $n_2$ = 1.33. Both condition #1 and #2 change. [0.2pt]

#1:

\phi_1 > \sin^{-1}\frac{1.33}{1.5} = 62.46^{\circ}

[0.2pt]

$\theta_1 < 90^{\mathrm{o}} - 62.46^{\mathrm{o}} = 27.54^{\mathrm{o}}$

Using Snell’s law,

\theta < \sin^{-1}\frac{1.5}{1.4}\sin(27.54^{\circ}) = 29.7^{\circ}

#2:

\theta_2 > \sin^{-1}\frac{1.33}{1.5} = 62.46^{\circ}

Using Snell’s law,

\theta > \sin^{-1}\frac{1.5}{1.4}\sin(62.46^{\circ}) = 71.81^{\circ}

[0.2pt]

As conditions #1 and #2 does not overlap. [0.1pt]

Again, light CANNOT be totally reflected back to the polymer. [0.2pt] [0,9 т.]

Part C: Light Propagation Through an Optical Fibre

C.1

Longitudinal cross section of an optical fibre: core n1 between two cladding layers n2, outside medium n3 on the left; a ray arrives at angle θa to the dashed z axis, is refracted into the core at angle θc and meets the core–cladding boundary at angle ϕc to the dashed normal; a blue dashed arrow shows the refracted ray running along the boundary. — Ray entering the fibre core and meeting the core/cladding boundary at the critical angle

In order for light to not refractive out from the fiber, the angle of refraction that the light makes with the normal of the fiber core/cladding boundary needs to be $90^{\mathrm{o}}$ . Using Snell’s law,

n_1\sin\phi_c = n_2\sin(90^{\circ}) \qquad (1)

[0.4pt]

Taking the square of eq. (1), we have,

n_1^2\sin^2\phi_c = n_2^2

Using the trigonometry identity, we have, [0.2pt]

n_1^2(1 - \cos^2\phi_c) = n_2^2

Rearranging, we got,

n_1^2\cos^2\phi_c = n_1^2 - n_2^2

Or, [0.2pt]

n_1^2\sin^2\theta_c = n_1^2 - n_2^2 \qquad (2)

Taking the square root of eq. (2), we have,

n_1\sin\theta_c = \sqrt{n_1^2 - n_2^2}

[0.4pt]

With this, we can use Snell’s law to calculate the maximum angle of light incident,

n_3\sin\theta_a = n_1\sin\theta_c = \sqrt{n_1^2 - n_2^2}

[0.4pt]

\theta_a = \sin^{-1}\frac{\sqrt{n_1^2 - n_2^2}}{n_3}

[0.4pt]

or $\theta_a = \sin^{-1}\sqrt{{n_1}^2 - {n_2}^2}$ [2 т.]

C.2

A fibre with core n1 of width 50 μm between cladding layers n2 that bends downward on the right about a centre C far below; a ray enters at angle θa′, travels at θc and ϕc′ in the straight section, reaches point B on the lower boundary (angle ϕc′ to the normal) and point A on the upper boundary of the bent section (angle ϕc to the normal), where part of the light leaves the core; blue dotted lines join C to B and C to A. — Bent fibre: ray path through the straight and the bent section

Identifying the correct relationship between $\phi_c$ and $\phi'_c$ [0.4pt]
Find the relationship between $\phi_c$ and $\phi_c$ ’:
Using the Law of Sine,

\sin\phi_c = \frac{CB}{CA}\sin(\pi - \phi_c')

[0.4pt]

For light guiding at point A,

\sin\phi_c = \frac{n_2}{n_1}

[0.4pt]

CB = 10000-25 = 9975 $\mu$ m

CA = 10000+25 = 10025 $\mu$ m [0.4pt]

\sin(\pi - \phi_c') = \sin(\phi_c')

Therefore,

\sin(\phi_c') = \frac{10025n_2}{9975n_1}

[0.4pt]

Rearrange into eq.(1) and derive following part (a):

a.n_1\sin(\phi_c') = n_2, \qquad where\ a = \frac{9975}{10025}

a^2\,n_1^2\sin^2(\phi_c') = n_2^2

a^2n_1^2(1 - \cos^2\phi'_c) = n_2^2

a^2n_1^2\cos^2\phi'_c = a^2n_1^2 - n_2^2

a^2n_1^2\sin^2\theta'_c = a^2n_1^2 - n_2^2

n_1\sin\theta'_c = \frac{1}{a}\sqrt{a^2n_1^2 - n_2^2}

[0.4pt]

To calculate the new angle of acceptance,

n_3\sin\theta'_a = n_1\sin\theta'_c = \frac{1}{a}\sqrt{a^2n_1^2 - n_2^2}

\theta'_a = \sin^{-1}\left(\frac{10025}{9975n_3}\sqrt{a^2n_1^2 - n_2^2}\right)

[0.2pt] [2,6 т.]

C.3 For case C.1: State that $n_3$ = 1;

\sqrt{n_1^2 - n_2^2} = \sqrt{1.45^2 - 1.44^2} = 0.17

[0.1pt]

\theta_a = \sin^{-1}0.17 = 9.79^{0}

[0.1pt]

For case C.2: Substitute,

\theta'_a = \sin^{-1}\left(\frac{10025}{9975}\sqrt{\left(\frac{9975}{10025}\right)^2 1.45^2 - 1.44^2}\right)

[0.1pt]

= 5.15^{\circ}

[0.1pt] [0,4 т.]

Оригинал в Архива: th.pdf · официални решения: th_sol.pdf