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APhO 2006, Теория — Задача 3D. Part D

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Всички материали от APhO 2006

7th Asian Physics Olympiad — Theoretical Question 3 · 2,5 т.

Условие

D. [2.5 points] Two thin lenses with lens powers $D_1$ and $D_2$ are located at distance $L = 25cm$ from each other, and their main optical axes coincide. This system creates a direct real image of the object, located at the main optical axis closer to lens $D_1$ , with the magnification $\Gamma' = 1$ . If the positions of the two lenses are exchanged, the system again produces a direct real image, with the magnification $\Gamma'' = 4$ .

D1) What are the types of the lenses? On the answer sheet you should mark the gathering lens as «+», and the diverging lens as «-».

D2) What is the difference between the lens powers $\Delta D = D_1 - D_2$ ?

Отговори

Решение

First one has to determine the types of the lenses. If both lenses are negative, one always obtains a direct imaginary image. If one lens is positive and the other is negative, three variants are possible: an inversed real image, a direct imaginary image or an inversed imaginary image, all contradicting the conditions of the problem. Only the last variant is left – two positive lenses. The first lens creates an inversed real image, and the second one inverts in once more, creating the direct real image. Using the lens equations, the magnifications of the lenses can be written as

\Gamma_1 = \frac{F_1}{d_1 - F_1}; \qquad \Gamma_2 = \frac{F_2}{d_2 - F_2}, \qquad (30)

where $d_1$ is the distance from the object to the first lens, $d_2 = L - f_1$ is the distance from the image of the first lens to the second lens, and $f_1$ is the distance from the first lens to the first image. The total magnification of the system is $\Gamma' = \Gamma_1\cdot\Gamma_2$ . Using the expression for $d_2$ , inverted magnification coefficient can be written as

\frac{1}{\Gamma'} = \frac{d_1\left[L - (F_1 + F_2)\right]}{F_1 F_2} - \frac{L}{F_2} + 1. \qquad (31)

One notices from this expression that if two lenses are exchanged, the first term stays invariant, and only the second term changes. Thus the expression for the inverted magnification in the second case is:

\frac{1}{\Gamma''} = \frac{d_1\left[L - (F_1 + F_2)\right]}{F_1 F_2} - \frac{L}{F_1} + 1. \qquad (32)

Subtracting these two formulas, we get:

\frac{1}{\Gamma'} - \frac{1}{\Gamma''} = L\left(\frac{1}{F_1} - \frac{1}{F_2}\right) = L(D_1 - D_2); \qquad (33)

D_1 - D_2 = \frac{1}{L}\left(\frac{1}{\Gamma'} - \frac{1}{\Gamma''}\right) = \frac{1}{0{,}25}\left(1 - \frac{1}{4}\right) = \frac{1}{0{,}25}\cdot\frac{3}{4} = 3\ \text{diopters}. \qquad (34)

Mark Distribution

No	Total Pt	Partial Pt	Contents
D	2.5	1.0	Signs of lenses
		0.5	Expression for the magnification coefficient
		1.0	Final answer for $D_1 - D_2$

Оригинал в Архива: 3.pdf · официални решения: 3_sol.pdf