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INPhO 2021, Теория — Задача 4

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Всички материали от INPhO 2021

Indian Olympiad Qualifier in Physics (IOQP) 2020-2021 — Part II: Indian National Physics Olympiad (INPhO) · 7 февруари 2021 г. · 12 т.

Указания

This booklet consists of 4 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.
Marks will be awarded on the basis of what you write on both the Summary Answer Sheet and the Detailed Answer Sheets in the Answer Booklet. Simple short answers and plots may be directly entered in the Summary Answer Sheet. Marks may be deducted for absence of detailed work in questions involving longer calculations.
Strike out any rough work that you do not want to be considered for evaluation. You may also use the space on the Question Paper for rough work – this will NOT be evaluated.
Non-programmable scientific calculators are allowed. Mobile phones cannot be used as calculators.
Last page of the question paper can be used for rough work.
Please submit the Answer Booklet at the end of the examination. You may retain the Question Paper.

Table of Constants

Quantity	Symbol	Value
Speed of light in vacuum	$c$	$3.00 \times 10^{8}\ \mathrm{m{\cdot}s^{-1}}$
Planck's constant	$h$	$6.63 \times 10^{-34}\ \mathrm{J{\cdot}s}$
Magnitude of electron charge	$e$	$1.60 \times 10^{-19}\ \mathrm{C}$
Rest mass of electron	$m_e$	$9.11 \times 10^{-31}\ \mathrm{kg}$
Value of $1/4\pi\epsilon_0$		$9.00 \times 10^{9}\ \mathrm{N{\cdot}m^{2}{\cdot}C^{-2}}$
Acceleration due to gravity	$g$	$9.81\ \mathrm{m{\cdot}s^{-2}}$

Note on solutions

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

Условие

Three thin convex lenses $\mathrm{L_1}$ , $\mathrm{L_2}$ , and $\mathrm{L_3}$ with focal lengths $f_1$ , $f_2$ , and $f_3$ , respectively, are arranged in order ( $\mathrm{L_1}$ followed by $\mathrm{L_2}$ , followed by $\mathrm{L_3}$ from left to right) with their principal axes coincident. The distance $d_{12}$ between $\mathrm{L_1}$ and $\mathrm{L_2}$ , and the distance $d_{23}$ between $\mathrm{L_2}$ and $\mathrm{L_3}$ are such that $d_{12} + d_{23} \geq f_1 + 4f_2 + f_3$ . If a parallel beam of light incident on $\mathrm{L_1}$ at a small angle to the principal axis remains parallel to itself when leaving the system after passing through $\mathrm{L_2}$ and $\mathrm{L_3}$ , draw the appropriate ray diagram and determine $d_{12}$ and $d_{23}$ in terms of $f_1$ , $f_2$ , and $f_3$ .

Отговори

Решение

Solution:

An incoming parallel beam falling on the thin lens $\mathrm{L_1}$ will converge to a certain point A on the focal plane of $\mathrm{L_1}$ . The point A serves as the point source for $\mathrm{L_2}$ whose image is formed on the other side of $\mathrm{L_2}$ at a certain point B. The line AB must intersect the principal axis at the pole $\mathrm{O_2}$ of $\mathrm{L_2}$ . For a parallel beam to emerge from $\mathrm{L_3}$ , B must lie in the focal plane of $\mathrm{L_3}$ . The necessary ray diagram is drawn below.

Ray diagram: lenses L1, L2, L3 on a common principal axis. A parallel beam inclined at angle α converges through L1 to point A in its focal plane (A1 on the axis below it, at distance f1 from L1). A dashed line from A through the pole O2 of L2 reaches point B below the axis (B1 on the axis above it), which lies at distance f3 in front of L3; after L3 the rays emerge parallel, again at angle α. Distances d12 (L1 to L2) and d23 (L2 to L3) are marked below.

Since $\alpha \approx 0$ , we make the approximations

\mathrm{AA_1} = f_1 \tan\alpha \approx f_1\alpha

\mathrm{BB_1} = f_2 \tan\alpha \approx f_2\alpha

From magnification formula for lens $\mathrm{L_2}$ ,

\frac{\mathrm{BB_1}}{\mathrm{AA_1}} = \frac{f_3\alpha}{f_1\alpha} = \frac{v_2}{-u_2} = \frac{d_{23} - f_3}{d_{12} - f_1}

\implies \frac{d_{12}}{f_1} = \frac{d_{23}}{f_3} = k(\text{say})

\implies d_{12} = kf_1 \quad \text{and} \quad d_{23} = kf_3

From the lens equation for lens $\mathrm{L_2}$ ,

\frac{1}{v_2} - \frac{1}{u_2} = \frac{1}{f_2}

\frac{1}{(d_{23} - f_3)} - \frac{1}{-(d_{12} - f_1)} = \frac{1}{f_2}

\frac{1}{f_3(k-1)} + \frac{1}{f_1(k-1)} = \frac{1}{f_2}

\implies \quad k = 1 + \frac{f_2}{f_1} + \frac{f_2}{f_3}

Then,

\boxed{d_{12} = f_1 + f_2 + \frac{f_1 f_2}{f_3}}

\boxed{d_{23} = f_2 + f_3 + \frac{f_2 f_3}{f_1}}

Оригинал в Архива: IOQP2021-PartII-Questions-en.pdf · официални решения: IOQP2021-PartII-Solutions-20210217.pdf