6. (15 points) Looking down from the edge of the stream bank, Vovochka decided that the height of his rubber boots would be enough to cross the stream. However, after crossing, Vovochka got his legs wet up to his knees ($H=52$ cm). Estimate the height $h$ of Vovochka's boots. Assume the depth of the stream is constant, and the refractive index of water $n=1.33$.

Rector of SPbPU

![](https://cdn.mathpix.com/cropped/2024_05_06_080bff219c908526ae49g-1.jpg?height=379&width=776&top_left_y=2502&top_left_x=699)

A.I. Rudskoy S.A. Staroytov

## Instructions for Evaluating Final Round Solutions 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 received for each individual problem.

The score for solving each problem in the version is one of the following coefficients:

1.0 - the problem is solved correctly;

0.8 - the problem is solved correctly and the answer is given 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 first column "For the Teacher" of the "Answer Sheet". In the second column of the section, the score is entered, which is the product of the coefficient received for the solution and the maximum score for the given problem.

The obtained scores are summed and entered in the "Total" row.

Table for converting coefficients to points.

| \multirow{coefficient}{max points}{} | 10 | 15 | 20 | 25 |
| :---: | :---: | :---: | :---: | :---: |
| 1 | 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 |
6. From the figure

$\frac{d}{h}=\operatorname{tg} \alpha ; \quad \frac{d}{H}=\operatorname{tg} \beta ; \quad \frac{H}{h}=\frac{\operatorname{tg} \alpha}{\operatorname{tg} \beta}$

Since all the information about the bottom of the stream falls into the space limited by the eye's pupil, all angles are small: $\operatorname{tg} \alpha \approx \sin \alpha ; \operatorname{tg} \beta \approx \sin \beta$

Then $\frac{H}{h}=\frac{\sin \alpha}{\sin \beta}=n$

$h=\frac{H}{n}=39 \mathrm{~cm}$.

(15 points)