Kvant Physics Problem 319

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Verdicts: PASS + PASS
Solve time: 9m12s
Source on kvant.digital

Problem

At a distance $2F$ from a converging lens $\text{Л}_1$ with focal length $F$, a luminous object is located. The illuminance of the sharp image of the object on a screen is equal to $E_0$. Between the screen and the lens $\text{Л}_1$, a diverging lens $\text{Л}_2$ with focal length $-2F$ is placed. To obtain a sharp image of the object, the screen had to be moved by a distance equal to $F$. Determine the illuminance of the image of the object in the second case.

V. E. Belonuchkin

Setup and Assumptions

A luminous object is placed on the optical axis of a thin converging lens $\text{Л}_1$ with focal length $F$. The object distance from the lens is $s_1 = 2F$. In the initial configuration a sharp image is formed on a screen, and the illuminance of this image is $E_0$.

A thin diverging lens $\text{Л}_2$ with focal length $f_2 = -2F$ is inserted between $\text{Л}_1$ and the screen. After inserting the second lens, the screen is displaced along the optical axis by a distance $F$ so that a new sharp image is formed. The task is to determine the illuminance $E$ of the image in this second configuration.

All lenses are assumed thin and ideal, paraxial approximation is valid, energy losses due to absorption and reflection are neglected, and the object is treated as a Lambertian emitter with fixed surface brightness so that image illuminance is determined by geometrical imaging and magnification only.

Physical Principles

For a thin lens the imaging relation is given by

$$\frac{1}{F} = \frac{1}{s} + \frac{1}{s'}.$$

The transverse magnification of a thin lens is

$$m = -\frac{s'}{s}.$$

For an ideal lossless optical system the total luminous flux is conserved, and the illuminance in the image plane is inversely proportional to the square of the transverse magnification,

$$E \propto \frac{1}{|m|^2}.$$

For a system of thin lenses in series, the total magnification is the product of individual magnifications,

$$m = m_1 m_2.$$

Derivation

In the initial configuration only the lens $\text{Л}_1$ acts. With $s_1 = 2F$, the lens equation gives

$$\frac{1}{F} = \frac{1}{2F} + \frac{1}{s_1'}.$$

Hence

$$\frac{1}{s_1'} = \frac{1}{F} - \frac{1}{2F} = \frac{1}{2F},$$

so the image distance is $s_1' = 2F$.

The magnification of the first lens is

$$m_1 = -\frac{s_1'}{s_1} = -\frac{2F}{2F} = -1.$$

The initial image has unit magnification in absolute value, so the reference illuminance is $E_0$.

After insertion of the diverging lens $\text{Л}_2$, the final sharp image is formed on a screen shifted by $F$ relative to the original position. The original image plane was at $2F$ from $\text{Л}_1$, therefore the final image position is at $3F$ from $\text{Л}_1$.

Let the position of $\text{Л}_2$ be at a distance $x$ from $\text{Л}_1$. The intermediate image produced by $\text{Л}_1$ at $2F$ acts as a virtual object for $\text{Л}_2$. The object distance for $\text{Л}_2$ is $s_2 = 2F - x$, and the image distance is $s_2' = 3F - x$. The lens equation for $\text{Л}_2$ is

$$\frac{1}{-2F} = \frac{1}{2F - x} + \frac{1}{3F - x}.$$

Introducing the dimensionless variable $a = x/F$, this becomes

$$-\frac{1}{2} = \frac{1}{2-a} + \frac{1}{3-a}.$$

Combining the right-hand side,

$$\frac{1}{2-a} + \frac{1}{3-a} = \frac{(3-a)+(2-a)}{(2-a)(3-a)} = \frac{5-2a}{(2-a)(3-a)}.$$

Thus

$$-\frac{1}{2}(2-a)(3-a) = 5-2a.$$

Expanding,

$$(2-a)(3-a) = 6 - 5a + a^2,$$

$$-\frac{1}{2}(a^2 - 5a + 6) = 5 - 2a.$$

Multiplying by $2$ gives

$$-(a^2 - 5a + 6) = 10 - 4a,$$

which leads to

$$-a^2 + 9a - 16 = 0.$$

Multiplying by $-1$,

$$a^2 - 9a + 16 = 0.$$

The physically admissible root lying between $0$ and $3$ is

$$a = \frac{9 - \sqrt{17}}{2}.$$

The magnification of the second lens is

$$m_2 = -\frac{s_2'}{s_2} = -\frac{3F - x}{2F - x}.$$

The total magnification is

$$m = m_1 m_2 = (-1)\left(-\frac{3F - x}{2F - x}\right) = \frac{3F - x}{2F - x}.$$

Substituting $x = aF$ gives

$$|m| = \frac{3 - a}{2 - a}.$$

Using the relation between illuminance and magnification,

$$E = E_0 \frac{1}{|m|^2} = E_0 \left(\frac{2-a}{3-a}\right)^2.$$

Substituting $a = \frac{9 - \sqrt{17}}{2}$,

$$2 - a = \frac{\sqrt{17} - 5}{2}, \qquad 3 - a = \frac{\sqrt{17} - 3}{2}.$$

Therefore,

$$E = E_0 \left(\frac{\sqrt{17} - 5}{\sqrt{17} - 3}\right)^2.$$

Numerically,

$$\sqrt{17} \approx 4.123,$$

$$E \approx E_0 \left(\frac{-0.877}{1.123}\right)^2 \approx 0.61,E_0.$$

Result

$$\boxed{E = E_0 \left(\frac{\sqrt{17} - 5}{\sqrt{17} - 3}\right)^2 \approx 0.61,E_0}$$

Sanity Checks

The expression for $E$ depends only on dimensionless ratios of distances measured in units of $F$, ensuring dimensional consistency. The magnification is close to unity in magnitude because both lenses operate with comparable focal lengths, which is consistent with the moderate change in illuminance.

The result $E < E_0$ follows from the fact that the second lens system increases the effective transverse magnification magnitude above unity, spreading the same luminous flux over a larger image area. The most sensitive step is the determination of the correct physical root for $x$, since the quadratic equation admits two mathematical solutions, and selecting the root outside the interval between lens and screen would correspond to an unphysical placement of the second lens relative to the formed image.