When is the distance between the object and the real image produced by a convex lens the smallest?
I. Solution. Speaking of a real image, the object and the image are on opposite sides of the lens. If the distances from the optical center of the lens are $t$ and $k$, then the quantity to be examined is

$$
y=t+k, \text { if } \frac{1}{t}+\frac{1}{k}=\frac{1}{f}
$$

From the latter law,

$$
k=\frac{t f}{t-f}
$$

and

$$
y=t+\frac{t f}{t-f}=\frac{t^{2}}{t-f}
$$

For a real image, $t \geq f$, so $y>0$.

If $t=f, y=+\infty$ and if $t=+\infty$, $y=+\infty$ as well.

Accordingly, $y$ is a single-valued, continuous function of $t$, which is positive everywhere in the interval in question; hence, it must have at least one minimum within that interval, and indeed at the point where its first derivative vanishes.

$$
y^{\prime}=\frac{d y}{d t}=\frac{2 t(t-f)-t^{2}}{(t-f)^{2}}=\frac{t^{2}-2 t f}{(t-f)^{2}}=0
$$

if

$$
t^{2}-2 t f=t(t-2 f)=0
$$

Thus, $y^{\prime}=0$ if $t=0$ or if $t=2 f$.

$t=0$ does not fall into the interval under consideration, only $t=2 f$.

If $t<2 f, y^{\prime}<0$ and if $t>2 f, y^{\prime}>0$, so the function $y$ indeed has a minimum at $t=2 f$ (it transitions from decreasing to increasing).

The smallest value of $y$ is: $4 f$. (That is, $t=2 f$ and $k=2 f$.)

Sebestyén Gyula (Fazekas Mihály r. VIII. o. Debrecen)

Note. $1^{0}$. Some of the solutions are satisfied with the statement at the extremum examination that "the minimum is where the first derivative is zero."

The correct statement is that an extremum can occur where $y^{\prime}=0$. Whether this extremum is a minimum or a maximum is primarily determined by the change in sign of $y^{\prime}$ or the sign of $y^{\prime \prime}$. However, in simpler cases, it is also possible to draw conclusions without this; for example, if the function is continuous, has a constant sign, is zero or infinite at the endpoints of the interval, and $y^{\prime}$ vanishes only at one point in the interval.

$2^{0}$. If $t<2 f$, then $y$ decreases as $t$ increases, and if $t>2 f$, $y$ increases as $t$ increases. (If $t>2 f$, the image distance decreases to a lesser extent.)

# II. Solution. From the equation

$$
\frac{1}{t}+\frac{1}{k}=\frac{1}{f}
$$

it follows that:

$$
t+k=\frac{k t}{f}=\frac{1}{f}: \frac{1}{k} \cdot \frac{1}{t}
$$

The value of $t+k$ is minimum when the product $\frac{1}{k} \cdot \frac{1}{t}$ is maximum. Since the sum of the factors of this product is constant, the product is maximum when the factors are equal, i.e., when $k=t=2 f$.

Kádár Géza (Dobó István r. VIII. o. Eger)