Съдържание

INPhO 2025, Теория — Задача 5. Metalens

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от INPhO 2025

Indian National Physics Olympiad (INPhO)-2025 — Question Paper · 2 февруари 2025 г. · 15 т.

Внимание: Бележка към темата

The complete retained original question and matching HBCSE solution packets are transcribed. The separately referenced Answer Booklet is not retained in the matching archive row. Printed key formula/sign/symbol/prose discrepancies and the original handwritten magnetic-data graph remain literal, with separately labelled source notes.

Original cover and instructions

Indian National Physics Olympiad (INPhO)-2025

HOMI BHABHA CENTRE FOR SCIENCE EDUCATION

Tata Institute of Fundamental Research

V. N. Purav Marg, Mankhurd, Mumbai, 400 088

Question Paper

Date: 02 February 2025

Time: 09:00-12:00 (3 hours)

Maximum Marks: 75

Instructions — Roll Number: $\square\square\square\square-\square\square\square\square-\square\square\square\square$

This booklet consists of 9 pages and total of 5 questions. Write roll number at the top wherever asked.
Booklet to write the answers is provided separately. Instructions to write the answers are on the Answer Booklet.
Non-programmable scientific calculators are allowed. Mobile phones cannot be used as calculators.
Please submit the Answer Sheet at the end of the examination. You may retain the Question Paper.

Question Number	1	2	3	4	5	Total
Maximum Marks	8	14	16	22	15	75

Table of Constants

Speed of light in vacuum	$c$	$3.00\times10^8\ \mathrm{m\cdot s^{-1}}$
Magnitude of electron charge	$e$	$1.60\times10^{-19}\ \mathrm C$
Value of $1/4\pi\epsilon_0$		$9.00\times10^9\ \mathrm{N\cdot m^2\cdot C^{-2}}$
Avogadro’s number	$N_A$	$6.022\times10^{23}\ \mathrm{mol^{-1}}$
Acceleration due to gravity	$g$	$9.81\ \mathrm{m\cdot s^{-2}}$
Universal Gas Constant	$R$	$8.31\ \mathrm{J\cdot K^{-1}\cdot mol^{-1}}$
	$R$	$0.0821\ \mathrm{l\cdot atm\cdot mol^{-1}\cdot K^{-1}}$
Boltzmann constant	$k_B$	$1.3806\times10^{-23}\ \mathrm{J\cdot K^{-1}}$
Permeability constant	$\mu_0$	$4\pi\times10^{-7}\ \mathrm{H\cdot m^{-1}}$

Native key notice

The matching key repeats the cover, constant table, marks table and four instructions with “24 pages” and the title “Solutions”.

Please note that alternate/equivalent methods and different ways of expressing final solutions may exist. A correct method will be suitably awarded.

Условие

A metasurface is a two-dimensional, ultra-thin optical structure consisting of an array of nanospaced optical nano-elements (also known as meta-atoms) on a flat surface (typically an ultra-thin glass plate). The primary function of the nano-elements is to locally introduce a phase shift $\phi(\vec r)$ to the wave, incident at position $\vec r$ . This function, $\phi(\vec r)$ , is called the phase profile of the metasurface.

Metalens has a circular metasurface with a circularly symmetric phase profile function, $\phi(r)$ , which depends on the distance $r$ of the point from the center of the metalens (see figure below). This type of metalens can be used for focusing incoming parallel rays to a point. Unlike normal lenses, the metalens will look just like an ultrathin circular disc.

Whole native circular metalens disc/x/y/z/r and Metalens arrow label.

Consider two homogeneous media of refractive indices $n_1$ and $n_2$ separated by a metalens as shown in the figure below. Suppose a beam of plane wave is incident at point $Q$ (at distance $r$ from the pole $P$ ) on the metasurface at an angle of $\theta_1$ from medium 1. Assume that the rays falling at $Q$ are in the plane containing $PQ$ and the axis of the lens ( $x$ - $z$ plane). These rays will be refracted at an angle $\theta_2$ in the medium 2. The angle of refraction $\theta_2$ depends on $r$ . Thus, the modified law of refraction for the metasurface can be written as

n_1\sin\theta_1-n_2\sin\theta_2=f(r)

Similarly, the rays falling at $Q'$ , at a distance $r'$ , will be refracted by an angle $\theta'_2$ .

Full metalens refraction geometry, both media/n1/n2/P/Q/Q-prime/r/r-prime/theta1/theta2/theta2-prime/axes.

a) Find $f(r)$ in terms of $\phi(r)$ and $k_o$ , the wave number of the incoming wave in a vacuum. To determine $f(r)$ , assume two rays in $x$ - $z$ plane incident at an angle $\theta_1$ at two infinitesimally close points, $r$ and $r+\Delta r$ , are refracted by the same angle $\theta_2$ . You don’t need to derive the exact functional form of $\phi(r)$ for this part. [8 т.]

b) Derive an expression for the phase profile $\phi(r)$ , to convert a plane wavefront to spherical wavefront, with light being focused to a point $F$ on the axis (see figure below), which is at a distance $f$ from the pole P.

Full parallel-ray focus geometry, n1/n2/P/F/f/axes and Metalens.

[4 т.]

c) Consider a metalens whose phase profile is obtained in part (b). For the paraxial approximation, derive an expression for the lens equation, having object distance $u$ and image distance being $v$ , with focal length $f$ (see figure below).

Full native object/image ray geometry with u/v/axes/n1/n2/P/Metalens.

[3 т.]

Решение

(a) Solution:

Entire native nearby-ray/wavefront diagram with OA/BB-prime/AB/A-prime-B-prime/theta1/theta2/r/Delta-r/xz geometry.

Consider two rays, which are close to each other, incident locally on metalens. The ray $OA$ is incident at an angle $\theta_1$ at $r$ and exits in medium 2 at an angle $\theta_2$ . Similarly the ray $BB'$ exits in medium 2 at an angle of $\theta_2$ from $r+\Delta r$ . The rays being very close to each other have a plane wavefront. Let wavefront AB propagate to A’B’ in time $t$ . Let $\psi_A,\psi_B,\psi_{A'}$ and $\psi_{B'}$ be phases at points $A,B,A'$ and $B'$ respectively. Then

\psi_{A'}-\psi_A=k_2\Delta r\sin\theta_2-\omega t+\phi(r)

\psi_{B'}-\psi_B=k_1\Delta r\sin\theta_1-\omega t+\phi(r+\Delta r)

Since these two phase differences must be equal, we get

k_2\Delta r\sin\theta_2-k_1\Delta r\sin\theta_1=\phi(r+\Delta r)-\phi(r)

n_2\sin\theta_2-n_1\sin\theta_1=\frac1{k_0}\frac{\phi(r+\Delta r)-\phi(r)}{\Delta r}\ \to\ \boxed{\frac1{k_0}\frac{d\phi(r)}{dr}}

Alternate solution:

Time interval for the wavefront to travel from $A$ to $A'$ is equal to the time taken by the wavefront to travel from $B$ to $B'$ .

\tau_{AA'}+\tau_{A'}=\tau_{BB'}+\tau_{B'}\qquad(5.1)

where $\tau_{BB'}$ is the time it takes for point B on the wavefront (traveling at speed $v_1$ ) to reach point B’, and $\tau_{AA'}$ is the time it takes the transmitted portion of that same wavefront (traveling at speed $v_2$ ) to reach point A’ from point A. The time delays due to the metasurface at points A and B’ are $\tau_A$ and $\tau_B$ , respectively. From figure

\tau_{AA'}=\frac1c\Delta r n_2\sin\theta_2

\tau_{BB'}=\frac1c\Delta r n_1\sin\theta_1

\tau_A=\frac{\phi_A}{ck_o}=\frac{\phi(r)}{ck_o}

\tau'_B=\frac{\phi_A}{ck_o}=\frac{\phi(r+\Delta r)}{ck_o}

Substituting above terms in Eq.5.1, we get

\Delta r\sin\theta_2+\frac{\Phi(r)}{k_o}=\Delta r\sin\theta_1+\frac{\Phi(r+\Delta r)}{k_o}

Simplifying:

\frac{\Phi(r+\Delta r)-\Phi(r)}{\Delta r}=\frac1{k_0}\frac{d\Phi}{dr}

This gives the relation:

\boxed{\frac1{k_0}\frac{d\Phi}{dr}=n_2\sin\theta_2-n_1\sin\theta_1}

(b) Solution: From the previous part

n_2\sin\theta_2-n_1\sin\theta_1=\frac1{k_o}\frac{d\phi}{dr}

From the above figure, the ray arriving at $r$ (with $\theta_1=0$ ) must bend by an angle $\theta_2=\sin^{-1}\left(\frac r{\sqrt{f^2+r^2}}\right)$ . Thus,

\frac{n_2r}{\sqrt{(f^2+r^2)}}-0=\frac1{k_o}\frac{d\phi}{dr}

Simplifying above equation, we get

\boxed{\phi(r)=\pm k_2\left(\sqrt{(r^2+f^2)}-f\right)}

where $k_2=k_0n_2$ .

(c) Solution: We know that

n_2\sin\theta_2-n_1\sin\theta_1=\frac1{k_o}\frac{d\phi}{dr}

For paraxial rays, we can use small angle approximation, $\sin\theta_1=r/u$ and then above equation becomes

n_2\theta_2-n_1\theta_1=\frac1{k_o}\frac{d\phi}{dr}

We know that

\frac1{k_0}\frac{d\phi}{dr}=\frac{n_2r}{\sqrt{(r^2+f^2)}}

For paraxial approximation, we get

\frac1{k_0}\frac{d\phi}{dr}=\frac{n_2r}{\sqrt{(r^2+f^2)}}\approx n_2\frac rf

Using these approximations, we get

\frac{n_1}{u}+\frac{n_2}{v}=\frac{n_2}{f}

If $n_1=n_2$ , then the above equation becomes

\boxed{\frac1u+\frac1v=\frac1f}\qquad(5.2)

Source note: The question defines $f(r)$ with $n_1\sin\theta_1-n_2\sin\theta_2$ , while the key derives the reverse order. The alternate solution prints $\tau_{A'},\tau_{B'}$ in (5.1), $\tau_A,\tau_B$ in its prose, and $\tau'_B=\phi_A/(ck_o)$ in its formulas. It changes lowercase $\phi$ to uppercase $\Phi$ , drops $n_1,n_2$ in the substitution line, and prints an additional $1/k_0$ on the finite-difference simplification. The source varies $k_o$ and $k_0$ and gives the phase profile with $\pm$ . All these native expressions, both alternative derivations and the complete paraxial key are retained literally.

Native source end

END OF THE QUESTION PAPER

The key also ends with this native wording on page24. Key page20 is blank apart from running page and roll-number furniture.

Оригинал в Архива: INPhO2025-Question.pdf · официални решения: INPhO2025-Solution-20250219.pdf