6. (20 points) A diverging lens produces an image of a nail that is reduced by a factor of 4. The nail is positioned on the principal optical axis of the lens with its head facing the lens. The length of the nail is $\ell=20$ cm, and the optical power of the lens is $D=-5$ diopters. Find the distance from the optical center of the lens to the head of the nail.

## Grading Guidelines for the Final Round of the Polytechnic Physics Olympiad

The version given to the participant contains 6 problems of varying difficulty.

The evaluation of the work is based on the points obtained for each individual problem.

For the results of the check, one of the following coefficients is assigned for solving each problem in the version:

1.0 - the problem is solved correctly;

0.8 - the problem is solved correctly and the answer is obtained in a general form; there is an error in the unit of measurement of the obtained physical quantity or an arithmetic error;

0.6 - the problem is not fully solved; all necessary physical relationships for solving the problem are present; there is an error in algebraic transformations

0.4 - the problem is not fully solved; some physical relationships necessary for solving the problem are missing;

0.2 - the problem is not solved; the work contains only some notes related to solving the problem or describing the phenomenon considered in the problem;

0.0 - the solution to the problem or any notes related to it are absent in the work.

The coefficient is entered in the "Column for the Instructor" of the "Answer Sheet" in the first column. In the second column of the column, the score is placed, which is equal to the product of the coefficient obtained for the solution and the maximum score for the given problem.

The obtained scores are summed and recorded in the "Total" line.

Table for converting coefficients to points.

| \multirow{coefficient}{max score}{} | 10 | 15 | 20 | 25 |
| :---: | :---: | :---: | :---: | :---: |
| 1.0 | 10 | 15 | 20 | 25 |
| 0.8 | 8 | 12 | 16 | 20 |
| 0.6 | 6 | 9 | 12 | 15 |
| 0.4 | 4 | 6 | 8 | 10 |
| 0.2 | 2 | 3 | 4 | 5 |
| 0.0 | 0 | 0 | 0 | 0 |
6. According to the problem:

$F=\frac{1}{D}=20 \mathrm{~cm}=\ell$

$d_{1}=d_{2}-F ; \quad f_{2}-f_{1}=\frac{1}{4} F$

From the thin lens formulas for the two ends of the nail, we find the distances to the ends of the image

$-\frac{1}{F}=\frac{1}{d_{2}-F}-\frac{1}{f_{1}} \rightarrow f_{1}=\frac{F\left(d_{2}-F\right)}{\left(d_{2}-F\right)+F}$

$-\frac{1}{F}=\frac{1}{d_{2}}-\frac{1}{f_{2}} \rightarrow f_{2}=\frac{F d_{2}}{d_{2}-F}$

![](https://cdn.mathpix.com/cropped/2024_05_06_8ff79e222c4198d5a7c2g-2.jpg?height=272&width=759&top_left_y=1920&top_left_x=1134)

$\frac{F d_{2}}{d_{2}-F}-\frac{F\left(d_{2}-F\right)}{d_{2}}=\frac{1}{4} F$

We obtain the quadratic equation $d_{2}^{2}+F d_{2}-4 F^{2}=0$

$d_{2}=F \frac{\sqrt{17}-1}{2} \quad d_{1}=\frac{\sqrt{17}-3}{2 D}=11.2$ cm.

(20 points)