Kvant Physics Problem 319

Verified: no
Verdicts: SKIP + SKIP
Solve time: 1m18s
Source on kvant.digital

Problem

At a distance $2F$ from a converging lens $\text{Л}_1$ with focal length $F$, there is a luminous object. The illuminance of the sharp image of the object on the screen in this case is $E_0$. Between the screen and the lens $\text{Л}_1$, a diverging lens $\text{Л}_2$ with focal length $-2F$ was 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

An object is placed on the optical axis of a thin converging lens $\text{Л}_1$ with focal length $F$, at a distance $2F$ from it. A sharp image is formed on a screen, and its illuminance is $E_0$.

A thin diverging lens $\text{Л}_2$ with focal length $-2F$ is inserted between $\text{Л}_1$ and the screen. After insertion, the screen is shifted by a distance $F$ to restore a sharp image.

The unknown is the new illuminance $E$ on the screen.

All lenses are assumed ideal and thin. The system is assumed to operate in the paraxial regime. Energy losses due to absorption and reflection are neglected. The object is assumed to be incoherently luminous so that illuminance is proportional to the optical flux per unit image area.

Physical Principles

The thin lens equation is

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

where $s$ is the object distance and $s'$ is the image distance.

For a lossless optical system, the total luminous flux through the system is conserved.

The illuminance at the image plane is proportional to the flux density, hence

$$E \propto \frac{\Phi}{A}$$

where $\Phi$ is the transmitted flux and $A$ is the image area.

For a given flux, the image area scales as the square of the transverse magnification $m$, so

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

The transverse magnification of a thin lens is

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

Derivation

For the first lens $\text{Л}_1$, the object is at $s = 2F$. The lens equation gives

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

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

hence

$$s_1' = 2F.$$

The first image is therefore formed at a distance $2F$ from $\text{Л}_1$, with transverse magnification

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

The insertion of $\text{Л}_2$ modifies the convergence of rays at the original image plane. At the plane where the intermediate image of $\text{Л}_1$ would have formed, the rays are converging toward a point, which is equivalent to a real object for $\text{Л}_2$ located at its front side.

Using vergence addition, the vergence of the beam at the plane of $\text{Л}_2$ is

$$V = -\frac{1}{2F}.$$

The optical power of $\text{Л}_2$ is

$$\Phi_2 = -\frac{1}{2F}.$$

After refraction, the outgoing vergence is

$$V' = V + \Phi_2 = -\frac{1}{2F} - \frac{1}{2F} = -\frac{1}{F}.$$

Thus the final image is formed at a distance

$$s_2' = F$$

to the right of $\text{Л}_2$.

Since $\text{Л}_2$ is placed at the former image region, the total distance from $\text{Л}_1$ to the final image is

$$s'_{\text{final}} = 2F + F = 3F.$$

The overall transverse magnification of the two-lens system is therefore

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

The image area scales as $m^2$, hence the ratio of illuminances is

$$\frac{E}{E_0} = \frac{m_1^2}{m_2^2} = \frac{1}{\left(\frac{3}{2}\right)^2}.$$

Result

$$E = E_0 \cdot \frac{4}{9}.$$

Substituting the given value,

$$\boxed{E = \frac{4}{9} E_0}$$

with $E$ measured in the same units as $E_0$.

Sanity Checks

The result is dimensionally consistent since the ratio $E/E_0$ is dimensionless.

The magnification increases in magnitude from $1$ to $3/2$, which increases the image area by a factor of $(3/2)^2$, so the illuminance must decrease by the same factor, matching $4/9$.

The only step sensitive to error is the interpretation of the intermediate vergence at the insertion plane of $\text{Л}_2$, since an incorrect sign in $V$ or $\Phi_2$ would invert the direction of image shift and change the effective magnification, leading to an incorrect brightness scaling.