Kvant Physics Problem 619

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Problem

Luminous point is located on the principal optical axis at a distance $d=40~\text{см}$ from a diverging lens with focal length $F=10~\text{см}$. The point is shifted by a distance $h=5~\text{см}$ in a plane perpendicular to the principal optical axis. By how much and in what direction should the lens be shifted so that the image of the luminous point returns to its original position?

Leningrad City Physics Olympiad (1979)

Setup and Assumptions

A thin diverging lens with focal length $F = -10~\text{cm}$ is fixed in space, and a luminous point source is initially located on the principal optical axis at distance $d = 40~\text{cm}$ from the lens. The image formed by the lens lies on the axis at some point determined by Gaussian imaging.

The source is then displaced perpendicular to the optical axis by a distance $h = 5~\text{cm}$ in a plane perpendicular to the axis. The lens is subsequently translated perpendicular to the axis by an unknown distance $x$, while remaining parallel to itself. The required condition is that the image of the displaced source returns to the original image position corresponding to the undeformed configuration.

The lens is assumed thin, aberration-free, and obeying paraxial optics. All displacements are small enough for linear ray optics to apply, and transverse shifts are treated within first-order approximation.

Physical Principles

The Gaussian lens formula relates object and image distances measured from the lens,

$$\frac{1}{F} = \frac{1}{v} + \frac{1}{u}.$$

For paraxial rays, transverse magnification is given by

$$m = \frac{y'}{y} = \frac{v}{u},$$

where $y$ and $y'$ are the transverse coordinates of the object and image measured in the lens reference frame.

A rigid transverse displacement of the lens by $x$ changes coordinates of both object and image in the lens frame, while leaving the direction of the principal axis fixed in space. Superposition holds for small transverse shifts, so contributions from object displacement and lens translation add linearly in the image coordinate.

Derivation

The initial axial position of the image is found from the lens formula. With $F = -10~\text{cm}$ and $u = 40~\text{cm}$,

$$\frac{1}{-10~\text{cm}} = \frac{1}{v} + \frac{1}{40~\text{cm}}.$$

This gives

$$-0.1~\text{cm}^{-1} = \frac{1}{v} + 0.025~\text{cm}^{-1},$$

$$\frac{1}{v} = -0.125~\text{cm}^{-1},$$

hence

$$v = -8~\text{cm}.$$

The transverse magnification is therefore

$$m = \frac{v}{u} = \frac{-8~\text{cm}}{40~\text{cm}} = -0.2.$$

When the object is shifted transversely by $h$, the lens is initially unshifted, so the image displacement relative to the axis is

$$y'_{(h)} = m h = (-0.2)(5~\text{cm}) = -1~\text{cm}.$$

Now the lens is translated transversely by $x$. In the lens frame, the object acquires an additional apparent displacement $-x$, so the lens translation contributes an image shift $m(-x)$ in the lens frame. Returning to the laboratory frame adds the rigid translation $x$, giving a net contribution

$$y'_{(x)} = x + m(-x) = x(1 - m).$$

By linear superposition of independent transverse shifts, the total image displacement is

$$y' = m h + x(1 - m).$$

The condition that the image returns to its original axial position is $y' = 0$, hence

$$m h + x(1 - m) = 0.$$

Solving for $x$ gives

$$x = -\frac{m h}{1 - m}.$$

Substituting $m = -0.2$ and $h = 5~\text{cm}$,

$$x = -\frac{(-0.2)(5~\text{cm})}{1 - (-0.2)}.$$

This yields

$$x = \frac{1~\text{cm}}{1.2} = 0.833~\text{cm}.$$

Result

The required lens displacement is

$$x = 0.833~\text{cm} \approx 8.3~\text{mm}.$$

The lens must be shifted in the same direction as the transverse displacement of the light source.

Sanity Checks

The expression $x = -\frac{m h}{1 - m}$ has dimensions of length since $m$ is dimensionless and $h$ carries units of length, ensuring dimensional consistency.

For a diverging lens, $m$ is negative, so the image of the displaced object shifts opposite to the object displacement, consistent with $y'_{(h)} = -1~\text{cm}$ for $h = 5~\text{cm}$. The required lens motion is smaller than the object displacement because the factor $1 - m = 1.2$ reduces the effectiveness of lens translation on the image position.

In the limiting case $m \to 0$, corresponding to a very weak imaging system, the formula gives $x \to 0$, since the image becomes insensitive to object displacement. In the opposite limit $m \to 1$, corresponding to unit magnification, the denominator tends to zero, indicating that lens translation alone cannot compensate object displacement, consistent with the geometry of rigid image transport.