Problem No. A1 (answer No. 4).
The angle of incidence of light on a horizontally located flat mirror equalsfont-size:14.0pt;line-height: 115%;font-family:" times new roman>. What will be the angle between the incident and reflected rays if you turn the mirror tofont-size:14.0pt;line-height: 115%;font-family:" times new roman> as shown in the picture? 1font-size:14.0pt;line-height: 115%;font-family:" times new roman>); 2)font-size:14.0pt;line-height: 115%;font-family:" times new roman>; 3)font-size:14.0pt;line-height: 115%;font-family:" times new roman>; 4)font-size:14.0pt;line-height: 115%;font-family:" times new roman>.
Given:
position:relative;top:5.5pt">font-size:14.0pt; line-height:115%;font-family:" times new roman>Find:
font-size:14.0pt; line-height:115%;font-family:" times new roman>Solution:
After rotating the mirror by 10°, the angle of incidence will be 20°, and the angle between the incident and reflected rays will be 40°.
Answer: the angle between the incident and reflected rays is 40°, therefore, answer No. 4.
Task No. A2. (answer #2)
An object located at double the focal length from a thin converging lens is moved to the focus of the lens. In this case, its image: 1) approaches the lens; 2) moves away from the focus of the lens; 3) approaches the focus of the lens; 4) approaches 2F.
Solution:
font-size:14.0pt;line-height:115%;font-family:" times new roman>As can be seen from the figure, if an object is moved towards the lens, then the image moves away from the focus of the lens.
Answer: moves away from the focus of the lens.
Problem No. A3 (answer No. 4).
Two sources emit electromagnetic waves frequencyfont-size:14.0pt;line-height: 115%;font-family:" times new roman> with the same initial phases. The minimum interference will be observed if the minimum difference in the wave paths is equal to: 1) 0; 2) 0.3 µm; 3) 0.6 µm; 4) 1 micron.
Given:
font-size:14.0pt; line-height:115%;font-family:" times new roman>Find:
font-size:14.0pt; line-height:115%;font-family:" times new roman>Solution:
The condition for the appearance of an interference minimum is determined by a very simple formula:
font-size:14.0pt; line-height:115%;font-family:" times new roman>Where font-size:14.0pt;line-height: 115%;font-family:" times new roman> - optical path difference, which must be found; m - maximum order (can take values 0, ±1, ±2, etc . etc.); λ - radiation wavelength. It remains to find out what exactly to substitute in the formula. m - can be selected at the end, looking at the answer options, but most likely we will need the first maximum m = 1. The wavelength is determined from the following dependence :
font-size:14.0pt; line-height:115%;font-family:" times new roman>Find the wavelength:
font-size:14.0pt; line-height:115%;font-family:" times new roman>from here
font-size:14.0pt; line-height:115%;font-family:" times new roman>Answer: font-size:14.0pt;line-height: 115%;font-family:" times new roman>.
Problem No. A4 (answer No. 2).
In experiments on the photoelectric effect, we took a metal plate with a work functionfont-size:14.0pt;line-height: 115%;font-family:" times new roman> and began to illuminate it with lightfont-size:14.0pt;line-height: 115%;font-family:" times new roman>. Then the frequency was reduced by 2 times, while simultaneously increasing the number of photons incident on the plate in 1 s by 1.5 times. As a result of this, the number of photoelectrons leaving the plate in 1 s: 1) increased by 1.5 times; 2) became equal to zero; 3) decreased by 2 times; 4) decreased by more than 2 times.
Given:
position:relative;top:5.5pt">position:relative;top:10.0pt">font-size:14.0pt; line-height:115%;font-family:" times new roman>Find:
font-size:14.0pt; line-height:115%;font-family:" times new roman>Solution:
Let's compare the energy of ejected electrons in the first and second cases. We use Einstein's equation for the photoelectric effect:
https://pandia.ru/text/79/268/images/image027_2.png" width="147" height="48 src=">
From the first equation, the energy of the ejected electrons is equal to
font-size:14.0pt; line-height:115%;font-family:" times new roman>The frequency is halved.
font-size:14.0pt; line-height:115%;font-family:" times new roman>Such light will not eject electrons, since the energy of ejected electrons cannot be less than zero. The photoelectric effect will stop.
Answer: the photoelectric effect will stop, i.e. the number of photoelectrons leaving the plate in 1 s has become zero.
Problem No. A5 (answer No. 1).
Which graph corresponds to the dependence of the maximum kinetic energy of photoelectrons E on the frequency ν of photons incident on the substance during the photoelectric effect (see figure)?
Font-size:14.0pt; line-height:115%;font-family:" times new roman>Solution:
Let's write Einstein's equation:
font-size:14.0pt;line-height: 115%;font-family:" times new roman> , at frequency equal to zero The work function is equal to the kinetic energy of the ejected electron:font-size:14.0pt;line-height: 115%;font-family:" times new roman> and the higher the frequency, the greater the kinetic energy.
It turns out that this is schedule No. 1.
Answer: schedule No. 1.
Problem No. A6 (answer No. 4).
The photon momentum has smallest value in the frequency range: 1) x-ray radiation; 2) visible radiation; 3) ultraviolet radiation; 4) infrared radiation.
Solution:
The photon momentum is calculated by the formula:
font-size:14.0pt; line-height:115%;font-family:" times new roman>where font-size:14.0pt;line-height: 115%;font-family:" times new roman> - radiation frequency, it is related to the wavelength by the ratio:font-size:14.0pt;line-height: 115%;font-family:" times new roman>. Infrared radiation has the longest wavelength, then its frequency is the smallest, therefore, the photon impulse has the smallest value in the frequency range of infrared radiation .
Answer: infrared radiation.
Problem No. A7 (answer No. 3).
How many photons of different frequencies can be emitted by hydrogen atoms in the second excited state?font-size:14.0pt;line-height: 115%;font-family:" times new novel>, according to Bohr's postulates?
Solution:
According to Bohr's postulate, when an electron transitions from one state (with higher energy) to another (with lower energy), a photon is emitted, thus, hydrogen atoms that were in the second excited statefont-size:14.0pt;line-height: 115%;font-family:" times new roman> electrons can transferfont-size:14.0pt;line-height: 115%;font-family:" times new roman>, thus, three photons of different frequencies can be emitted.
Answer: 3 photons.
Problem No. A8 (answer 1).
The particle speed isfont-size:14.0pt;line-height: 115%;font-family:" times new roman>. Its kinetic energy is 1)font-size:14.0pt;line-height: 115%;font-family:" times new roman>; 2)font-size:14.0pt;line-height: 115%;font-family:" times new roman>; 3)font-size:14.0pt;line-height: 115%;font-family:" times new roman>; 4)font-size:14.0pt;line-height: 115%;font-family:" times new roman>.
Solution:
At speeds close to the speed of light, the kinetic energy of any object is equal to
font-size:14.0pt; line-height:115%;font-family:" times new roman>Answer: the kinetic energy of the particle is font-size:14.0pt;line-height:115%; font-family:" times new roman>Task No. A9 (answer No. 4).
Radioactive isotopefont-size:14.0pt;line-height: 115%;font-family:" times new roman> after onefont-size:14.0pt;line-height: 115%;font-family:" times new roman> - collapse and twofont-size:14.0pt;line-height: 115%;font-family:" times new roman> - decays turns into an isotope of 1) protactinium; 2) uranium; 3) thorium; 4) radium.
Solution:
Let us write down the equations of decomposition reactions:
Alpha decay:
As a result of onefont-size:14.0pt;line-height: 115%;font-family:" times new roman> decay produces a thorium atom.
As a result of twofont-size:14.0pt;line-height: 115%;font-family:" times new roman> decays, a radium atom is formed.
Answer: radium atom
Problem No. A10 (answer 3).
Half-life of some radioactive isotope 1 month. How long will it take for the number of isotope nuclei to decrease by 32 times? 1) 3 months 2) 4 months 3) 5 months 4) 6 months.
Solution:
Let font-size:14.0pt;line-height: 115%;font-family:" times new roman>isotope activity in starting moment time,font-size:14.0pt;line-height: 115%;font-family:" times new roman>isotope activity through t months. Let's write down the law of decreasing activity:font-size:14.0pt;line-height: 115%;font-family:" times new roman>. divide the right and left parts intofont-size:14.0pt;line-height: 115%;font-family:" times new roman>:font-size:14.0pt;line-height: 115%;font-family:" times new roman> orfont-size:14.0pt;line-height: 115%;font-family:" times new roman>. Let's logarithm both parts:font-size:14.0pt;line-height: 115%;font-family:" times new roman> orfont-size:14.0pt;line-height: 115%;font-family:" times new roman>, from herefont-size:14.0pt; line-height:115%;font-family:" times new roman>Answer: the number of isotope nuclei will decrease by 32 times in 5 months.
Task No. B1.
A 2 m long fluorescent lamp is attached to the ceiling of a room 4 m high. At a height of 2 m from the floor, parallel to it there is a round opaque disk with a diameter of 2 m. The center of the lamp and the center of the disk lie on the same vertical. Find the minimum linear size of the shadow.
Solution:
font-size:14.0pt;line-height:115%;font-family:" times new roman>From the figure you can see that the minimum linear size of the shadow will coincide with the length of the lamp and the diameter of the disk and is equal to 2 m.
Task No. B2.
On diffraction grating, having a periodfont-size:14.0pt;line-height: 115%;font-family:" times new roman>, a normally parallel beam of white light is incident. The spectrum is observed on a screen located at a distance of 2 m from the grating. What is the distance between red and violet sections of the spectrum of the first order (the first color strip on the screen), if the wavelengths of red and purple light respectively equalfont-size:14.0pt;line-height: 115%;font-family:" times new roman> andfont-size:14.0pt;line-height: 115%;font-family:" times new roman>? Countfont-size:14.0pt;line-height: 115%;font-family:" times new roman>. Express your answer in cm.
Given:
position:relative;top:5.5pt">font-size:14.0pt;line-height: 115%;font-family:" times new roman>
position:relative;top:5.5pt">font-size:14.0pt;line-height: 115%;font-family:" times new roman>
Find:
font-size:14.0pt; line-height:115%;font-family:" times new roman>Solution:
Spectrum length font-size:14.0pt;line-height: 115%;font-family:" times new roman> we will find by subtracting from the distancefont-size:14.0pt;line-height: 115%;font-family:" times new roman> distance between the red line of the first order spectrum and the central maximumfont-size:14.0pt;line-height: 115%;font-family:" times new roman>between the violet spectrum band of the same order and the central maximum:font-size:14.0pt;line-height: 115%;font-family:" times new roman>. (1)
font-size:14.0pt;line-height:115%;font-family:" times new roman>According to the problem conditions font-size:14.0pt;line-height: 115%;font-family:" times new roman>, then from the figure it is clear thatfont-size:14.0pt;line-height: 115%;font-family:" times new roman>,
from here font-size:14.0pt;line-height: 115%;font-family:" times new roman>, (2)
We can similarly findfont-size:14.0pt;line-height: 115%;font-family:" times new roman>. (3)
From the condition of the maximum on the diffraction gratingfont-size:14.0pt;line-height: 115%;font-family:" times new roman> andfont-size:14.0pt;line-height: 115%;font-family:" times new roman>, wherefont-size:14.0pt;line-height: 115%;font-family:" times new roman>grid period, from where
font-size:14.0pt;line-height: 115%;font-family:" times new roman> andfont-size:14.0pt;line-height: 115%;font-family:" times new roman> . Let's substitute these expressions into formula (2) and (3) respectively.font-size:14.0pt;line-height: 115%;font-family:" times new roman> andfont-size:14.0pt;line-height: 115%;font-family:" times new roman> we substitute the resulting equations into formula (1) and the problem will be solved:.
Answer: the distance between the red and violet parts of the first order spectrum is 4 cm.
Task No. B3.
The photocathode is irradiated with light of wavelengthfont-size:14.0pt;line-height: 115%;font-family:" times new roman>. Red border of the photoelectric effect for the photocathode substancefont-size:14.0pt;line-height: 115%;font-family:" times new roman>. What voltage must be created between the anode and cathode for the photocurrent to stop?
Given:
position:relative;top:5.5pt">font-size:14.0pt;line-height:115%;font-family: " times new roman>Find:
font-size:14.0pt;line-height:115%;font-family: "times new roman>Solution:
Let's write down the expression for the photoelectric effect
Where
From here the voltage will be equal
Answer: 1.38 V.
Task No. B4.
A thin film of thickness is applied to the surface of the glass platefont-size:14.0pt;line-height: 115%;font-family:" times new roman>. Light of wavelength falls normally on the filmfont-size:14.0pt;line-height: 115%;font-family:" times new roman>. At what value of the refractive index of the film will the maximum reflection of light be observed?
Given:
font-size:14.0pt;line-height: 115%;font-family:" times new roman>
font-size:14.0pt;line-height: 115%;font-family:" times new roman>
Find:
font-size:14.0pt; line-height:115%;font-family:" times new roman>Solution:
Maximum illumination observed on the surface thin film in reflected light, meets the condition:
font-size:14.0pt;line-height: 115%;font-family:" times new roman>.
Here font-size:14.0pt;line-height: 115%;font-family:" times new roman>refraction angle. Because with normal incidence of rays on the filmfont-size:14.0pt;line-height: 115%;font-family:" times new roman>, andfont-size:14.0pt;line-height: 115%;font-family:" times new roman>, then you can rewritefont-size:14.0pt;line-height: 115%;font-family:" times new roman>, from herefont-size:14.0pt; line-height:115%;font-family:" times new roman>Answer: the refractive index of the film is equal tofont-size:14.0pt;line-height: 115%;font-family:" times new roman>.
Task No. B5.
What energy is released during a nuclear reaction?font-size:14.0pt;line-height: 115%;font-family:" times new roman> Express your answer in picojoules (pJ) and round to the nearest whole number.
Solution:
The amount of energy released during the reaction can be found using the formula:
Answer: 8pJ of energy is absorbed.
Task No. C1.
A pile hidden under water is driven vertically into the bottom of a reservoir 3 m deep. Pile height 2 m. Incidence angle sun rays to the surface of the water is equal tofont-size:14.0pt;line-height: 115%;font-family:" times new roman>. Determine the length of the shadow of the pile at the bottom of the reservoir. Refractive index of waterfont-size:14.0pt;line-height: 115%;font-family:" times new roman>.
Given:
font-size:14.0pt;line-height: 115%;font-family:" times new roman>
font-size:14.0pt;line-height: 115%;font-family:" times new roman>
font-size:14.0pt;line-height: 115%;font-family:" times new roman>
font-size:14.0pt;line-height: 115%;font-family:" times new roman>
Find:
font-size:14.0pt; line-height:115%;font-family:" times new roman>Solution:
font-size:14.0pt;line-height:115%;font-family:" times new roman>From the figure it is clear that the segments font-size:14.0pt;line-height: 115%;font-family:" times new roman> andfont-size:14.0pt;line-height: 115%;font-family:" times new roman> are the legs of a right triangle in which the angle is knownfont-size:14.0pt;line-height: 115%;font-family:" times new roman>, since triangle ABC similar to a triangle FEC , therefore, the anglefont-size:14.0pt;line-height: 115%;font-family:" times new roman>. Thenfont-size:14.0pt;line-height: 115%;font-family:" times new roman>, from here we express the desired valuefont-size:14.0pt;line-height: 115%;font-family:" times new roman>. Unknown anglefont-size:14.0pt;line-height: 115%;font-family:" times new roman>we find from the law of refraction:font-size:14.0pt;line-height: 115%;font-family:" times new roman>, from wherefont-size:14.0pt;line-height: 115%;font-family:" times new roman>.
We also do not know the angle of incidence of the beam on the surface of the water, but we know the height of the Sun above the horizonfont-size:14.0pt;line-height: 115%;font-family:" times new roman>, so the angle of incidencefont-size:14.0pt;line-height: 115%;font-family:" times new roman>we will find from equalityfont-size:14.0pt;line-height: 115%;font-family:" times new roman>, from herefont-size:14.0pt;line-height: 115%;font-family:" times new roman> andfont-size:14.0pt;line-height: 115%;font-family:" times new roman>, thenfont-size:14.0pt;line-height: 115%;font-family:" times new roman>. Now let's expressfont-size:14.0pt;line-height: 115%;font-family:" times new roman> viafont-size:14.0pt;line-height: 115%;font-family:" times new roman>. From the definition of tangent we havefont-size:14.0pt;line-height: 115%;font-family:" times new roman> orfont-size:14.0pt;line-height: 115%;font-family:" times new roman>. Thenfont-size:14.0pt;line-height: 115%;font-family:" times new roman>.
Answer: the length of the shadow of the pile at the bottom of the reservoir is 1.7 m.
Task No. C2.
font-size: 14.0pt;line-height:115%;font-family:" times new roman>Isosceles right triangle ABC area 50 font-size:14.0pt;line-height: 115%;font-family:" times new roman> is located in front of a thin converging lens so that its side AC lies on the main optical axis of the lens. The focal length of the lens is 50 cm. Top right angle C lies closer to the center of the lens than the vertex acute angle A. Distance from the center of the lens to point C (figure). Construct an image of a triangle and find the area of the resulting figure.
Given:
font-size:14.0pt;line-height: 115%;font-family:" times new roman>
font-size:14.0pt;line-height: 115%;font-family:" times new roman>
Find:
font-size:14.0pt; line-height:115%;font-family:" times new roman>Solution:
Let's make a drawing. It shows that the distance from side BC to the lensfont-size:14.0pt;line-height: 115%;font-family:" times new roman>, and the distance from the lens to the imagefont-size:14.0pt;line-height: 115%;font-family:" times new roman>, then using the formula
converging lensfont-size:14.0pt;line-height: 115%;font-family:" times new roman> can be written:font-size:14.0pt;line-height: 115%;font-family:" times new roman>, from herefont-size:14.0pt;line-height: 115%;font-family:" times new roman>, thenfont-size:14.0pt;line-height: 115%;font-family:" times new roman>.
Now let's look at the AC side; it lies on the main optical axis of the lens. Distancefont-size:14.0pt;line-height: 115%;font-family:" times new roman> determined from the lens formula:font-size:14.0pt;line-height: 115%;font-family:" times new roman>, wherefont-size:14.0pt;line-height: 115%;font-family:" times new roman>, thenfont-size:14.0pt;line-height: 115%;font-family:" times new roman> orfont-size:14.0pt;line-height: 115%;font-family:" times new roman>.
According to the conditions of the problemfont-size:14.0pt;line-height: 115%;font-family:" times new roman>, on the other handfont-size:14.0pt;line-height: 115%;font-family:" times new roman> can be rewrittenfont-size:14.0pt;line-height: 115%;font-family:" times new roman> orfont-size:14.0pt;line-height: 115%;font-family:" times new roman>.
Now you can findfont-size:14.0pt;line-height: 115%;font-family:" times new roman> :font-size:14.0pt;line-height: 115%;font-family:" times new roman>, thenfont-size:14.0pt; line-height:115%;font-family:" times new roman>Side of a trianglefont-size:14.0pt;line-height: 115%;font-family:" times new roman> equalfont-size:14.0pt;line-height: 115%;font-family:" times new roman>. DIV_ADBLOCK56">
Answer: the area of the resulting figure is equal tofont-size:14.0pt;line-height: 115%;font-family:" times new roman>.
Task No. C3.
Calcium coated photocathode (work functionfont-size:14.0pt;line-height: 115%;font-family:" times new roman>), illuminated with light frequencyfont-size:14.0pt;line-height: 115%;font-family:" times new roman>. Electrons emitted from the cathode enter a uniform magnetic field perpendicular to the induction lines and move in a circle of maximum radiusfont-size:14.0pt;line-height: 115%;font-family:" times new roman>. Calculate the induction modulus magnetic field IN.
Given:
position:relative;top:5.5pt">position:relative;top:5.5pt">font-size:14.0pt; line-height:115%;font-family:" times new roman>Find:
font-size:14.0pt;line-height: 115%;font-family:" times new roman>
Solution:
Let's use Einstein's formula for the photoelectric effect, according to which the photon energyfont-size:14.0pt;line-height: 115%;font-family:" times new roman> falling on the metal is spent on the work function of the electron from the metalfont-size:14.0pt;line-height: 115%;font-family:" times new roman> and to communicate kinetic energy to the torn electronfont-size:14.0pt;line-height: 115%;font-family:" times new roman>:font-size:14.0pt;line-height: 115%;font-family:" times new roman>, wherefont-size:14.0pt;line-height: 115%;font-family:" times new roman>, thenfont-size:14.0pt;line-height: 115%;font-family:" times new roman> orfont-size:14.0pt;line-height: 115%;font-family:" times new roman>, from herefont-size:14.0pt;line-height: 115%;font-family:" times new roman>. (1)
An electron moving in a magnetic field is acted upon by the Lorentz forcefont-size:14.0pt;line-height: 115%;font-family:" times new roman>, equal to Newton’s second law:font-size:14.0pt;line-height: 115%;font-family:" times new roman>, wherefont-size:14.0pt;line-height: 115%;font-family:" times new roman>, thereforefont-size:14.0pt;line-height: 115%;font-family:" times new roman>. (2)
According to the Lorentz force formulafont-size:14.0pt;line-height: 115%;font-family:" times new roman>, wherefont-size:14.0pt;line-height: 115%;font-family:" times new roman> andfont-size:14.0pt;line-height: 115%;font-family:" times new roman> thereforefont-size:14.0pt;line-height: 115%;font-family:" times new roman> (3)
The left sides of equations (2) andfont-size:14.0pt;line-height: 115%;font-family:" times new roman> are equal, therefore the right sides are equal:font-size:14.0pt;line-height: 115%;font-family:" times new roman>, orfont-size:14.0pt;line-height: 115%;font-family:" times new roman> from here
Answer: The magnetic field induction is 1.6 mT.
Problem No. C4.
Weight elementary particle equal tofont-size:14.0pt;line-height: 115%;font-family:" times new roman>, own time life is equalfont-size:14.0pt;line-height: 115%;font-family:" times new roman>. Which the path will pass during its lifetime, this particle, if its energy is equal to E?
Enter path font-size:14.0pt;line-height: 115%;font-family:" times new roman> accurate to an integer for timefont-size:14.0pt;line-height: 115%;font-family:" times new roman> and energyfont-size:14.0pt;line-height: 115%;font-family:" times new roman>.
Given:
position:relative;top:5.5pt">font-size:14.0pt;line-height:115%;font-family: " times new roman>Find:
font-size:14.0pt;line-height:115%;font-family: "times new roman>Solution:
Let us write a formula relating the coordinate time of a particlefont-size:14.0pt;line-height: 115%;font-family:" times new roman>, during which it travels the distancefont-size:14.0pt;line-height: 115%;font-family:" times new roman> with speedfont-size:14.0pt;line-height: 115%;font-family:" times new novel>, with her own time:position:relative;top:23.5pt">font-size:14.0pt;line-height: 115%;font-family:" times new roman>, wherefont-size:14.0pt;line-height: 115%;font-family:" times new roman>, thenfont-size:14.0pt;line-height: 115%;font-family:" times new roman>, from herefont-size:14.0pt;line-height: 115%;font-family:" times new roman> (1)
Let's find the speed font-size:14.0pt;line-height: 115%;font-family:" times new roman>. In relativistic mechanics, kinetic energy is calculated as the difference between total energy and rest energy:font-size:14.0pt;line-height: 115%;font-family:" times new roman>. According to the condition givenfont-size:14.0pt;line-height: 115%;font-family:" times new roman>,font-size:14.0pt;line-height: 115%;font-family:" times new roman>,font-size:14.0pt;line-height: 115%;font-family:" times new roman>, then taking into account the relativistic formula for kinetic energy we have:
font-size:14.0pt;line-height: 115%;font-family:" times new roman> orfont-size:14.0pt;line-height: 115%;font-family:" times new roman> orfont-size:14.0pt;line-height: 115%;font-family:" times new roman> orfont-size:14.0pt;line-height: 115%;font-family:" times new roman> orfont-size:14.0pt;line-height: 115%;font-family:" times new roman> orposition:relative;top:10.0pt">font-size:14.0pt;line-height:115%;font-family: " times new roman>Then we substitute the resulting expression into formula (1) and find the path:
Answer: 52 m.
Task No. C5.
At font-size:14.0pt;line-height: 115%;font-family:" times new roman>in the decay of a resting plutonium-239 nucleus, the mass defect (the difference in the mass of the reaction products and the mass of the initial nucleus) isfont-size:14.0pt;line-height: 115%;font-family:" times new roman> Find the speed and kinetic energy of the resulting nucleus. Massposition:relative;top:5.5pt">font-size:14.0pt; line-height:115%;font-family:" times new roman>Given:
position:relative;top:5.5pt">position:relative;top:5.5pt">position:relative;top:5.5pt">font-size:14.0pt; line-height:115%;font-family:" times new novel>Find:
font-size:14.0pt; line-height:115%;font-family:" times new roman>Solution:
The mass of the nucleus at rest is equal to (1)
Let's write down the law of conservation of energyfont-size:14.0pt;line-height: 115%;font-family:" times new roman> and impulsefont-size:14.0pt;line-height: 115%;font-family:" times new roman> systems of these particles. Since the particle was at rest before decay, its total momentumfont-size:14.0pt;line-height: 115%;font-family:" times new roman> and it should remain the same after the breakup:font-size:14.0pt;line-height: 115%;font-family:" times new roman>, i.e. the particles after the decay stage move antidirectionally, thereforefont-size:14.0pt; line-height:115%;font-family:" times new roman>According to the law of conservation of energy, the particle’s own energyfont-size:14.0pt;line-height: 115%;font-family:" times new roman> before the collapse is equal to the amount total energy one particlefont-size:14.0pt;line-height: 115%;font-family:" times new roman> and full of energy anotherfont-size:14.0pt;line-height: 115%;font-family:" times new roman>:font-size:14.0pt;line-height: 115%;font-family:" times new roman>, wherefont-size:14.0pt;line-height: 115%;font-family:" times new roman>, thereforefont-size:14.0pt;line-height: 115%;font-family:" times new roman>. (1)
Now let's connect the energy of each particle with its momentum:font-size:14.0pt;line-height: 115%;font-family:" times new roman>, from wherefont-size:14.0pt;line-height: 115%;font-family:" times new roman>,
And font-size:14.0pt;line-height: 115%;font-family:" times new roman> , sincefont-size:14.0pt;line-height: 115%;font-family:" times new roman>, thenfont-size:14.0pt;line-height: 115%;font-family:" times new roman>, from wherefont-size:14.0pt;line-height: 115%;font-family:" times new roman>. Thenfont-size:14.0pt;line-height: 115%;font-family:" times new roman> (2)
Let us express from equation (1) the energyfont-size:14.0pt;line-height: 115%;font-family:" times new roman> and put it in formula (2):
font-size:14.0pt;line-height: 115%;font-family:" times new roman>, thenfont-size:14.0pt;line-height: 115%;font-family:" times new roman> or
font-size:14.0pt;line-height: 115%;font-family:" times new roman> orfont-size:14.0pt;line-height: 115%;font-family:" times new roman> orfont-size:14.0pt;line-height: 115%;font-family:" times new roman>, from here
Total particle energyfont-size:14.0pt;line-height: 115%;font-family:" times new roman> is equal to the sum of its own energyfont-size:14.0pt;line-height: 115%;font-family:" times new roman> and kinetic energyfont-size:14.0pt;line-height: 115%;font-family:" times new roman>:font-size:14.0pt;line-height: 115%;font-family:" times new roman>, from herefont-size:14.0pt;line-height: 115%;font-family:" times new roman> orfont-size:14.0pt;line-height: 115%;font-family:" times new roman> orfont-size:14.0pt;line-height: 115%;font-family:" times new roman>, from herefont-size:14.0pt;line-height: 115%;font-family:" times new roman>.
Answer: the speed of the formed nucleus isfont-size:14.0pt;line-height: 115%;font-family:" times new roman> and energy -font-size:14.0pt;line-height: 115%;font-family:" times new roman>.
M. Yu. Demidova,
, FIPI, Moscow;
G. G. Nikiforov,
, ISMO RAO, FIPI, Moscow
Main results of the Unified State Exam-2007 in physics
Main results of the Unified State Exam-2007 in physicsM.Yu.DEMIDOVA, [email protected] ,
G.G.NIKIFOROV, Moscow
Main results of the Unified State Exam-2007 in physics
Analysis of completing tasks with detailed answers (part 3 of Unified State Exam tasks)
Traditionally, calculation problems were used high level complexity (level of entrance exams to universities) in four sections school course physics. On average, 63% of the total number of test takers started to complete them. Unlike Part 1, which is within the capabilities of everyone, these problems are solved mainly by students who receive grades “4” and “5” on the Unified State Examination. Therefore, the influence of difficulties in mathematics on both the choice of problems and their solutions is much less noticeable here. Preference is given to problems with standard formulations, with a clear physical situation, although they require a rather labor-intensive solution of a system of equations. Preference is given to problems in mechanics on the laws of conservation of energy and momentum (63% begin to solve), on the application of the first law of thermodynamics to isoprocesses (46%), as well as on the application of the laws direct current(45%). Tasks in the new situation turn out to be significantly less attractive. For example, only 16% of those tested dared to start doing problems involving electron diffraction on a crystal.
The highest results were shown when solving problems in mechanics, MCT and thermodynamics, although with an unconventional formulation. For example:
An inclined plane intersects a horizontal plane in a straight line AB. Angle between planes = 30°. A small puck begins to move up an inclined plane from a point A With initial speed 0 = 2 m/s at an angle = 60° to the straight line AB. During the movement, the puck slides onto a straight line AB at the point B. Neglecting the friction between the washer and inclined plane, find the distance AB.
The sample solution to this problem, given in the materials for experts, was performed based on kinematics equations, but many graduates used the law of conservation of energy to solve the problem. Unfortunately, when performing these tasks, errors in geometric transformations and recording projections of vectors onto selected axes, which significantly influenced the overall results.
Among the tasks on electrostatics and direct current, difficulties were caused by the task of oscillating a charged ball above a charged plane. At the same time, on average, 32% of those tested began to perform it, but 17% received 0 points (i.e., they could not understand how the charged plane affects the oscillations of the pendulum), 9% were able to write down the equation for the period of oscillation of the pendulum and the force acting on charge in an electrostatic field, receiving 1 point for this, and only 6% were able, with varying degrees of success, to bring these transformations to the answer and scored 2–3 points.
As in previous years, optics problems turned out to be the least attractive for graduates. Moreover, two related to well-known subjects (interference and image in a lens), and one was formulated in such a way that it was necessary to first independently isolate the phenomenon of total internal reflection. All had approximately the same volume and complexity of mathematical transformations. The low performance results were quite planned for a full-scale task. internal reflection, but for the problem geometric optics(see example) turned out to be much lower: 
Isosceles right triangle ABC with an area of 50 cm2 is located in front of a thin collecting lens so that its leg A.C. lies on the main optical axis of the lens. Lens focal length 50 cm. Right angle vertex C lies closer to the center of the lens than the vertex of the acute angle A. Distance from the center of the lens to the point C equal to twice the focal length of the lens. Construct an image of a triangle and find the area of the resulting figure.
A little more than half of the students chose this series of problems, but 27% were unable to cope with constructing the image of a triangle in a lens and received 0 points. The stumbling block turned out to be the construction of an image of the peak A. It can be assumed that the reduction in hours spent studying physics at school does not make it possible, even when preparing for the Unified State Exam, to show techniques for constructing images in a lens of points lying on its main optical axis, using a secondary optical axis. As a rule, after correctly constructing the image of a triangle, applying the lens formula and finding the area of the image triangle did not cause significant difficulties. But only 8% of those tested managed to bring the solution to the “ideal”.
For the second year in a row, the variants included interference problems in a thin wedge. Thus, in 2006, only 3% of those tested coped with the task of light interference in a wedge-shaped soap film. This year, 4% completed a similar interference task in a wedge of glass plates. It must be said that the old plot did not greatly affect the popularity of the task: last year about 15% chose it, this year about 18%, with the same results. Since to solve the problem it is necessary to know only the condition for observing the maximum (minimum) of interference and the relationship for a right triangle, such a low result is not caused by mathematical difficulties, but by the lack of experience in solving problems of this type.
It should be noted that the tasks described above (on constructing an image in a lens, oscillations of a charged pendulum in an electrostatic field and the interference of light in a wedge) are typical only for specialized study of physics, since only in this case there is a sufficient amount of study time. Those who have studied a physics course at a basic level find themselves in a situation of almost complete novelty: they know the basic laws, but situations are completely new, for example, constructing images of objects lying on the optical axis in a lens, or the nature of interference in a wedge. 
Among the tasks in quantum physics, the most difficult problems turned out to be the application of the laws of conservation of energy in the interaction of an atom with an electron. For example:
Suppose that the diagram of energy levels of atoms of a certain substance has the form shown in the figure, and the atoms are in a state with energy E(1) . An electron colliding with one of these atoms, as a result, acquired some additional energy. The electron momentum after a collision with an atom at rest turned out to be 1.2 × 10 –24 kg m/s. Determine the kinetic energy of the electron before the collision. Neglect the possibility of an atom emitting light upon collision with an electron.
Only 26% started solving this problem, of which 9% were able to write down individual elements of the solution (change in the kinetic energy of the electron and the transition of the atom to another energy level or the relationship between momentum and kinetic energy) and only 3% of graduates were able to provide a complete correct solution this task.
In some problems, it was necessary to independently propose a physical model, since there was no explicit description of it in the text: oscillation of a load on a spring connected through a fixed block to a block sliding on the table; electron diffraction on a crystal, the movement of a charged particle in a circle under the influence of the Coulomb force. The objective complexity of the novelty of the situation significantly influences the results (feasibility 4–7%). For example: 
The figure shows a diagram of a device for preliminary selection of charged particles for the purpose of their subsequent detailed study. The device is a capacitor, the plates of which are bent by an arc with a radius R 50 cm. Let us assume that ions with a charge – e, as it shown on the picture. Tension electric field in the capacitor the absolute value is 50 kV/m. The ion speed is 2 · 10 5 m/s. Ions with what mass will fly through the capacitor without touching its plates? Assume that the distance between the plates of the capacitor is small, the electric field strength in the capacitor is the same in absolute value everywhere, and there is no electric field outside the capacitor. Neglect the influence of gravity.
These problems can be solved practically “in one formula”; you just need to understand that centripetal acceleration is created by the Coulomb force. As a rule, problems with an implicitly specified physical model selects the smallest number of test takers, and this type of task is characterized by a different distribution of the average percentage of the primary scores from standard tasks. For example, for standard task in mechanics for the application of the laws of conservation of energy and momentum, the distribution is as follows: 0 points - 19%, 1 point - 16%, 2 points - 7%, 3 points - 9%. It is clear that more than half of the test takers “recognized” the condition of the problem, and of those who also knew how to apply conservation laws to inelastic impact, some made mistakes in writing the basic equations, and some could not cope with mathematical difficulties.
For a problem with a non-standard formulation of the condition, the distribution of points looks somewhat different. For example, for the above problem, 6% scored 0 points, 3% scored 1 point, 2% scored 2 points, and 13% scored 3 points. That is, few dared to start solving the problem, but those who understood the processes described were able to bring the solution to a successful end almost “without losses.”
Dynamics of completion of Unified State Exam tasks in 2002–2007. By certain species activities
In KIMs in physics in 2007, up to 40% of the tasks from previous years were used. Comparison of the results of these groups of tasks in Unified State Exam of different years allows us to identify the dynamics of mastering individual types of activities and content elements.
About 20% of tasks basic level in part 1 are aimed at checking the recognition of various laws and formulas, as well as their application for analysis complex processes at a quality level. There is a positive trend in the quality of their implementation, which suggests that the main “list of formulas” of the physics course at the reproductive level is being absorbed quite well. The only exception this year became a formula for the dependence of the capacitance of a flat-plate capacitor on the area of the plates and the distance between the plates.
For calculation tasks to check the same list of laws and formulas, the performance is either the same or slightly reduced. For example:
Hydrostatic pressure, baseline: 2005 – 68%, 2007 – 68%.
Water is poured into a vessel 20 cm deep, the level of which is 2 cm below the upper edge of the vessel. What is the pressure of the water column on the bottom?
1) 2 10 5 Pa; 2) 2000 Pa; 3) 1800 Pa; 4) 180 Pa.
Mechanical work, basic level: 2003 - 69%, 2007 - 63%.
A boy is carrying his friend on a sled along a horizontal road, applying a force of 60 N. The speed of the sled is constant. The sled rope makes an angle of 30° with the horizontal. On some part of the way mechanical work The elastic force of the rope is 6000 J. What is the length of this section of the path?
Most likely, this is due to problems with computational skills, mainly due to weak students.
Some increase in performance is also observed for the most complex type of activity tested - problem solving, which takes up more than 30% of the total volume of the exam version, and both for individual tasks higher level with a choice of answer and with a short answer, and for complex tasks with a detailed answer. Select any general trends difficult, because here the tasks are updated annually (only 1-2 old tasks). For example:
Law of conservation of energy, increased level: 2005 – 37%, 2007 – 41%.
A car moving with the engine turned off had a speed of 30 m/s on a horizontal section of the road. How far will he travel up the mountain slope at an angle of 30° to the horizon until his speed decreases to 20 m/s? Ignore friction.
1) 12.5 m; 2) 25 m; 3) 50 m; 4) 100 m.
Formula thin lens, increased level: 2004 – 59%, 2007 – 65%.
The focal length of a converging lens is 60 cm. At what distance from the lens is the virtual image of an object located at a distance of 40 cm from the lens? Express your answer in centimeters (cm). ( Answer: 120.)
Unfortunately, as in 2006, the most problematic are the qualitative questions that test understanding of the meaning of various concepts, quantities and laws, and control the ability to explain physical phenomena, highlight the conditions for their occurrence or distinguish the manifestations of these phenomena in surrounding life. They are characterized not only by quite low level fulfillment and lack positive dynamics, but sometimes a decrease in results. 
Uniformly accelerated motion, increased level: 2003 - 52%, 2005 - 54%, 2007 - 54%.
The figure on the right shows a graph of the velocity of a body versus time at straight motion. Which of the graphs expresses the dependence of the modulus of the resultant of all forces acting on the body on the time of movement? The reference system is considered inertial.
Postulates of SRT, basic level: 2003 – 52%, 2007 – 40%.
Which of the following statements are postulates? special theory relativity?
A) The principle of relativity – the equality of all inertial reference systems;
B) invariance of the speed of light in vacuum - the invariance of its modulus when moving from one inertial system countdown to another.
1) Only A; 2) only B; 3) both A and B; 4) neither A nor B.
The most clear tendency for a slight decrease in the quality of assignments is manifested for those questions for which the study time is primarily reduced with a general decrease in the number teaching hours allocated for the teaching of physics (for example, properties of electromagnetic waves, elements of special relativity, current in various media, elements of geometric optics, etc.).
Analysis of the results of work performed by students with different levels of training
The basic level tasks of Part 1 make it possible to clearly identify students with an unsatisfactory level of preparation (exam mark “2”); advanced level tasks ( A 8, A 9, A 15, A 23, A 24, IN 1–IN 4) – differentiate between “excellent” students, “good” students and “C” students, and tasks of a high level of complexity ( WITH 1–WITH 6) – highlight the “excellent students”. 
At unsatisfactory level preparation, test takers show an extremely low level of knowledge of even the basic conceptual apparatus of a school physics course: completion rate is 29% for multiple-choice tasks; 5% for short answer items; 0% for tasks with a detailed answer (i.e. fragmentary knowledge of individual formulas and some phenomena). For example, for the given graph of the dependence of the coordinates of a body on time when it moves from a point A (X= 0) to point IN (X= 30 km) and back speed on the section AB finds 70% of students in this group, and at the site VA– only 45%. The standard task is to calculate the elastic force of a spring by its stiffness k= ... N/m stretching by ... m is performed by 73%, and elongation in the reverse task is performed by only 48%.
At satisfactory level training demonstrated knowledge of the basic laws and formulas of the school physics course: 51% completion rate for multiple-choice tasks; 18% for short answer items; 4% for tasks with a detailed answer. There is no big gap when performing similar tasks to check laws and formulas, i.e. the impact of mathematical difficulties in simple calculations is much less than in the previous group. For example, for a series of tasks on friction force (calculation of the coefficient of friction, body weight, friction force, normal pressure force during uniform movement), the completion rate is 55–67%. Almost all learned content elements represent the laws and formulas that are most “worked through” in lessons. However, only 50% cope with questions of a qualitative nature (left-hand rule, arrangement of ranges of electromagnetic radiation by increasing or decreasing wavelength or frequency, patterns of phenomena electromagnetic induction and so on.). This group tries to begin solving individual problems with short and detailed answers, but students cannot complete the solution, as a rule, correctly writing down the condition of the problem and individual equations for solving it. The priority here are tasks in mechanics on the laws of conservation of energy and momentum, in electrodynamics on writing Ohm's law for a complete circuit and formulas for the Lorentz force.
At good level preparation Test takers show systematic knowledge of the school physics course when completing tasks basic and increased levels difficulties: 74% for multiple choice items; 48% for short answer items; 22% for tasks with a detailed answer. They cope with most questions of a qualitative nature, checking the peculiarities of the occurrence of phenomena. (Only two series of tasks at the basic level turned out to be too complex - the completion rate was less than 50%: the behavior of a dielectric in an electric field and the change in the number of photoelectrons when the energy of the incident light changes.) This group quite successfully completes a number of tasks at an advanced level, both with a choice of answers and with short answer (20–75%). Compared to last year, the results of completing tasks of a high level of complexity have improved. As a rule, test takers begin solving 3–4 problems with detailed answers, correctly write down basic laws and formulas, and do not experience serious mathematical difficulties when solving systems of equations. Therefore, they usually manage to bring solutions to problems in mechanics to the correct answer ( WITH 1), molecular physics (WITH 2) or electrodynamics. However, they prefer either not to begin solving problems with a non-standard formulation (for example, on electron diffraction or the interaction of electrons with an atom), or give up solving them halfway due to a lack of understanding of the processes described.
At excellent level preparation the average performance of tasks is as follows: 88% for tasks with a choice of answers; 77% for short answer items; 62% for tasks with a detailed answer. Typically, they complete (above 60%) an average of four out of six tasks in Part 3. Compared to the previous group, they not only have a wealth of knowledge, but have fully mastered it and can operate freely conceptual apparatus school physics course, understand the peculiarities of rather complex processes and phenomena. This is especially true for problems with non-traditional formulations, when solving which it is necessary to imagine the processes and phenomena taking place, and not just remember those discussed in class.
In conclusion, we give examples of typical tasks and how they can be performed by groups with different backgrounds. 
A basic level of. The figure shows the location of two fixed point electric charges +2q and + q and three points are indicated A, IN And WITH. The modulus of the intensity vector of the total electric field of these charges has:
1) highest value at the point A;
2) the largest value at a point IN;
3) the highest value at a point WITH;
4) same values at all three points.
Feasibility
Increased level. An electromagnetic wave from a certain source propagates in benzene, with a wavelength of 1.2 mm. Determine the oscillation period of the source. The refractive index of benzene is 1.5. Express your answer in picoseconds (10–12 s). ( Answer: 6.)
Feasibility
High level. A hollow metal ball weighing 2 g is suspended on a silk thread and placed above a positively charged plane creating a uniform vertical electric field with a strength of 10 6 V/m. The ball has positive charge 10 –8 Cl. The period of small oscillations of the ball is 1 s. What is the length of the thread? Round the answer in centimeters (cm) to whole numbers. ( Answer. 38.)
Feasibility
The proposed manual is part of the educational and methodological complex “Physics. Preparation for the Unified State Exam” and is intended for high-quality preparation for the physics exam. IN last years the passing score required for admission to prestigious universities. This leads to the need to have sufficiently high Unified State Examination points, including physics. To do this, when preparing for the exam, you should pay attention to Special attention for solving part C tasks. The manual is addressed primarily to those graduates who plan to get the maximum possible results at the Unified State Examination in Physics high score. It can provide significant assistance to methodologists and teachers in preparing graduates for the final certification.
Examples.
An isosceles right triangle ABC with an area of 50 cm2 is located in front of a thin converging lens so that its leg AC lies on the main optical axis of the lens. The focal length of the lens is 50 cm. The vertex of the right angle C lies closer to the center of the lens than the vertex of the acute angle A. The distance from the center of the lens to point C is equal to twice the focal length of the lens (see Fig. 34). Construct an image of a triangle and find the area of the resulting figure.
A balloon, the shell of which has a mass M = 145 kg and a volume V = 230 m3, is filled with hot air at normal atmospheric pressure and ambient air temperature t0 = 0°C. What minimum temperature t must the air inside the shell have for the ball to begin to rise? The shell of the ball is inextensible and has a small hole in the lower part.
- Physics, Unified State Exam, All sections of the course, Theory, tasks of basic and advanced levels of complexity, Monastyrsky L.M., 2016
- Unified State Exam 2016, Physics, 25 training options, Monastyrsky L.M., Bogatin A.S., Bezuglova G.S., 2015
- Physics, solution book, preparation for the Unified State Exam-2015, book 2, educational manual, Monastyrsky L.M., Bogatin A.S., Bogatina V.N., Bezuglova G.S., Beilin V.A., Ignatova Yu. A., Kolesnik D.V., Shevtsov V.A., 2015
- Physics, Preparation for the Unified State Exam-2015, Book 2, Monastyrsky L.M., Bogatin A.S., 2014
The following textbooks and books:
- Physics, Typical mistakes and complex topics on the Unified State Exam (Part C), Monastyrsky L.M., 2013
- Physics, Thematic tests for preparation for the Unified State Exam, Tasks of a high level of complexity, C1-C6, Monastyrsky L.M., Bogatin L.M., Ignatova Yu.A., 2013
Answer: 1,47 .
№13.2 On the table there is a vessel with a mirror bottom and matte walls. Falls to the bottom of an empty vessel a ray of light, as it shown on the picture. On the wallCDIn this case, a “bunny” can be observed in the vessel - the glare of the reflected beam. A certain amount of water is poured into the vessel. How does the height of the “bunny” point change?? Neglect the reflection of the beam from the surface of the liquid.
Answer: "Bunny" will climb up the wall CD . The height will increase.
№13.3. On a sunny day, the height of the shadow from a vertically placed meter ruler is 50 cm, and from a tree - 6 m. What is the height of the tree?
Answer:12 m.
№13.4. A 1 m high peg placed vertically close to street lamp, casts a shadow 0.8 m long. If you move the peg 1 m further from the lantern, it will cast a shadow 1.25 m long. At what height is the lantern suspended?
Answer: 3.2 m.
№13.5. Electric lamp placed in a frosted glass ball with a radius of 20 cm and suspended at a height of 5 m above the floor. A ball with a radius of 10 cm is held under a lamp at a height of 1 m from the floor. Find the radii of the shadow and penumbra cast by the ball. The symmetry axes of the ball and the ball coincide.
Answer: 7.5 cm: 17.5 cm.
№13.6. Light falls on a plane mirror. The angle between the incident ray and the reflected ray is 40° . Why equal to the angle between the incident beam and the mirror?
Answer:70 ° .
№13.7. The beam hits the mirror perpendicularly. At what angle will the reflected beam deviate from the incident beam if the mirror is rotated by 15° ?
Answer: 30 ° .
№13.8. The boy moves towards a plane mirror. Over some time, he approached the mirror by 40 cm. By how many meters during the same time will the distance between the boy and his image in the mirror decrease?
Answer: 0.8 m.
№13.9. A round pool with a radius of 5 m is filled to the brim with water. A lamp hangs above the center of the pool at a height of 3 m from the surface of the water. How far from the edge of the pool can a 1.8 m tall person move and still see the reflection of the lamp in the water?
Answer:3m.
№13.10. At what angle will a ray of light deviate from its original direction when falling at an angle of 45° onto the surface of the glass? On the surface of the diamond?
Answer: 19., 28˚.
№13.11. To a diver underwater, the sun's rays appear to fall at an angle of 60° to the surface of the water. What is the angular altitude of the Sun above the horizon?
Answer: Approximately 48˚.
№13.1 2. A ray of light hits the surface of the water at an angle of 40°. At what angle must the beam hit the glass surface for the angle of refraction to be the same?
Answer:Approximately 50˚.
№13.1 3. A boy tries to hit an object with a stick at the bottom of a stream 40 cm deep. At what distance from the object will the stick hit the bottom of the stream if the boy, with precise aim, moves the stick at an angle of 45° to the surface of the water?
Answer: 15 cm.
№13.14. The figure on the left shows two flat mirrors (Z1 and M2) and a beam horizontally incident on mirror 1. Mirror 2 is rotated relative to the horizontal axis passing through point O at an angle of 15° (figure on the right). What is the angle between the rays reflected from mirror 1 and from mirror 2?
Answer:30 °
№13.15 The angle of incidence of light on a horizontally located plane mirror is 30°. What will be the angle between the incident and reflected rays if the mirror is rotated 10° as shown in the figure?
Answer: 40˚.
№13.16 The refracted ray of light makes an angle of 90 with the reflected ray° . Find relative indicator refraction if the beam falls on a flat interface between two media at an angle of 53° .
Answer: 1,3 .
№13.17 A pile hidden under water is driven vertically into the bottom of a reservoir 3 m deep. The height of the pile above the bottom is 2 m. The angle of incidence of sunlight on the water surface is 30°. Determine the length of the shadow of the pile at the bottom of the reservoir. Refractive index of water.
Answer: Approximately 0.8 m.
№13.18 A rectangular inflatable raft 6 m long floats on the surface of the water. The sky is covered with a continuous cloud cover, completely dispersing sunlight. The depth of the shadow under the raft is 2.3 m. Determine the width of the raft. Neglect the immersion depth of the raft and the scattering of light by water. The refractive index of water relative to air is assumed to be equal to .
Answer: Approximately 5.2 m.