Test No. 5

"Light Phenomena"

Option 1

Level A

1. An example of a phenomenon that proves the rectilinear propagation of light could be

formation of a wake in the sky from a jet aircraft

existence of tree shadow

mirage over the desert

unchanged position North Star in the sky

2. A ray of light falls on a flat mirror. The reflection angle is 24°. Angle between incident beam and mirror

1) 12° 2) 102° 3) 24° 4) 66°

3. A person who was at a distance of 4 m from a flat mirror moved and found himself at a distance of 3 m from the mirror. How much did the distance between the person and his image change?

1) 6 m 2) 4 m 3) 2 m 4) 1 m

4. If an object is located from a collecting lens at a distance greater than double the focal length (see figure), then its image is

real, inverted and augmented

real, direct and augmented

imaginary, inverted and reduced

real, inverted and diminished

5. The man wears glasses, focal length which is equal to 50 cm. Optical power the lenses of these glasses are equal to

D = 2 diopters 3) D = 0.02 diopters

D = - 2 diopters 4) D= - 0.02 diopters

6. To obtain a clear image on the retina of the eye, when moving the gaze from distant objects to close ones, it changes

shape of the lens 3) shape of the eyeball

pupil size 4) fundus shape

Level B

7. Establish a correspondence between light sources and their nature.

LIGHT SOURCES

A) Lightning

B) Fireflies

B) Comet

For each position in the first column, select the corresponding position in the second and write down to the table selected numbers under the corresponding letters.

THEIR NATURE

Thermal

Reflective light

Gas discharge

Luminescent

Level C

8. Construct an image of a luminous point after passing through the lens system.

Key to the tasks of option 1

Option 2

Level A

1. The shadow on the screen from an object illuminated by a point light source is 3 times larger than the object itself. The distance from the light source to the object is 1 m. Determine the distance from the light source to the screen.

1) 1 m 2) 2 m 3) 3 m 4) 4 m

2. A ray of light falls on a flat mirror. The angle of incidence was reduced by 5°. Angle between a plane mirror and the reflected ray

increased by 10° 3) decreased by 10°

increased by 5° 4) decreased by 5°

3. A person moves away from a plane mirror. His image in the mirror

remains in place 3) moves away from the mirror

approaches the mirror 4) becomes blurry

4. What will be the image of an object in a converging lens if the object is between the focus and the double focus of the lens?

Real, inverted and magnified

Real, direct and augmented

Imaginary, inverted and reduced

Real, inverted and diminished

5. What is the optical power of a diverging lens if its focal length is (- 10 cm)?

0.1 diopters 3) - 10 diopters

OD diopters 4) + 10 diopters

6. The boy wears glasses with diverging lenses. What is his vision defect?

Farsightedness 3) Myopia

Colorblindness 4) Astigmatism

Level B

7. Establish a correspondence between optical instruments and the basic physical phenomena underlying the principle of their operation.

For each position in the first column, select the corresponding position in the second and write down to the table selected numbers under the corresponding letters.

A) Periscope

B) Projector

B) Camera

PHYSICAL PHENOMENON

2) Light reflection

3) Refraction of light

4) Light scattering

Level C

8. Construct an image of a luminous point after passing through the lens system.

Laboratory work No. 1

Study of changes in temperature of cooling water over time.

Goals: investigate the change in temperature of cooling water over time

Devices and materials: vessel with hot water(70 o C – 80 o C), glass, thermometer

Exercise:

Determine the thermometer division price

Pour hot water weighing 100 - 150 g into a glass

Place the thermometer in the water and take its reading every minute; enter the measurement results in the table

Based on the data obtained, construct a graph of temperature changes over time, while marking the time on the OX axis t, and along the OU axis – temperature T

Compare the changes in water temperature that occurred during one of the first and one of the last minutes of the cooling process. Draw a conclusion about whether the water cools evenly in areas of higher and higher low temperatures. At what temperatures does water cool faster?

Results

General lesson on the topic “Light phenomena”

(8th grade)

The mind is not only about knowledge,

but also in the ability to apply knowledge in practice.
Aristotle.

Lesson type: lesson of repetition and correction of knowledge

Objective of the lesson: generalize and deepen students’ knowledge on the topic “Light Phenomena”, develop the ability to apply them when explaining experiments, physical phenomena at school, on the street, at home.

Lesson objectives:

EDUCATIONAL:

direct students' activities to

generalization and deepening of knowledge on the topic “Light phenomena”;

application theoretical knowledge to solve various problems;

EDUCATIONAL:

fostering an attitude towards physics as a science necessary for understanding the processes occurring in the world around us;

fostering cooperation in the process of jointly performing tasks;

DEVELOPMENTAL:

developing interest in the subject and the need to deepen and expand knowledge;

development of students’ thinking, memory, independence, hard work; creativity students.

development of intellectual and cognitive abilities students.

Equipment:

computer

TV (multimedia projector)

painting by I. Shishkin “Morning in pine forest”, painting by V. Polenov “Overgrown Pond”, photograph of a sunset near a pond.

Lesson plan.

Lesson progress

Organizational moment

Dear colleagues, students, I am glad to welcome you to the physics lesson.

The word “physics” comes from the Greek word “fusis”, which means... (nature) Slide 1

The word (physics) appeared for the first time in the works greatest thinker antiquity (Aristotle). Slide 2

Thanks to which scientist did the word (physics) enter the Russian language?

The word (physics) entered the Russian language thanks to a Russian scientist. Who is this great scientist you found out at home by answering the question

He was the first to construct optical instruments for astronomical observations and discovered the atmosphere of Venus. (L O M O N O S O V) Slide 3 (speaker with presentation)

In the previous lesson, we finished studying the last chapter in the 8th grade physics course. What was the name of this chapter?

We will write a test for the next lesson. What do you need to write it successfully?

Setting goals and objectives. (Students formulate goals and objectives themselves)

Updating of reference knowledge

Intelligence consists not only in knowledge, but also in the ability to apply knowledge in practice.
(Aristotle) ​​Slide 4

Let's remember the basic concepts, phenomena and laws of the chapter, watch the video.

Checking homework

1. Light falls on the mirror. The angle between the incident and reflected rays is 50°. At what angle does the light fall to the mirror? Draw it in your notebook.

Now let’s work with the drawing problems. You must choose the correct answer and, at my command, show the signal card.

Where will the light ray hit?

2. Indicate in what case is a lens hidden in a box divergent?

From a number of words you need to choose a concept from the topic “Light phenomena”

Fizminutka

“Overgrown Pond” by Vasily Dmitrievich Polenov.

It seems to me, guys, that there are a lot of fish in this pond. Let's try to catch her with a spear.

Why, while in a boat, is it difficult to hit a fish swimming nearby with a spear?

(The image of a fish in water is imaginary, raised to the surface, because water is a medium optically denser than air. Therefore, you should aim with the spear so that there is a gap between the fish and the tip.)

We sit on the shore together,
We look into a transparent body of water.
A grain of sand on the bottom is visible.
Tell me: what is the depth?
“It’s up to my neck here,” you said.
Jump down, but didn’t reach the bottom.
Here he emerged from under the water...
But why were you wrong?

Why do you think this happened?

Summarizing material on the topic

Demonstration: a ball, thickly covered with soot, is dropped into a glass of water. The ball appears shiny. When we take the ball out of the water, we see that it is covered with soot. Why did the ball in the water seem shiny?

(Soot particles are poorly wetted by water. Therefore, an air film forms around the soot ball. Due to the complete reflection of light from the layer of air formed between the soot and water, the ball appears shiny.)

Constructions

By completing these tasks, we will repeat with you the laws of propagation, reflection and refraction of light, the construction of images in a flat mirror and images given by lenses.

Let's think about interesting facts

Work in groups (teams). In a team, everyone has their own responsibilities, which they strive to fulfill conscientiously, so as not to let their comrades down. But in case of difficulty, each member of the group can count on the help of their friends. Now each team (the class is divided into groups of four) will be offered several tasks. Your task is to complete them efficiently and with the least amount of time. Each team member performs one of the tasks. In case of difficulty, you are allowed to ask your comrades for help.

1. How is a rainbow formed?

Rainbow

We all admire the rainbow - one of the most beautiful natural phenomena. The rainbow has been poeticized by many peoples. The Slavs believed that during a thunderstorm, the thunder god strikes evil spirits with lightning. A rainbow appearing after a rain and thunderstorm meant, in their opinion, triumph good forces who have conquered evil.

I. Newton, who studied spectra, identified seven colors in the rainbow, although, of course, this is rather arbitrary.

To explain the appearance of a rainbow, we will use a video recording

2. “Morning in a pine forest” by Ivan Ivanovich Shishkin.

Deaf pine forest awakens. The sun had just risen and flooded everything around with its generous rays. Take a look at sun rays.

What do we call a ray of light in optics? And what natural phenomena help us see the sun's rays in the picture? How does this happen?

In experiments to study light phenomena, we use narrow beams of light obtained through various holes. Is it possible to obtain a light beam by reducing the diameter of the hole?

(In fog, light is scattered (reflected) by small droplets of water.)

3. Guys, many of you have probably read “Mysterious island” J. Verna. Remember?

“But who lit the fire? - asked the sailor.

“The sun,” Spilett replied...

Indeed, the Sun delivered the fire that the sailor admired so much. He couldn’t believe his eyes and was so amazed that he couldn’t even question the engineer.

So you had a burning glass? - Herbert asked the engineer.

No, I made it.

And he showed. These were simply two glasses removed by the engineer from his watch and Spilett's watch. He connected their edges with clay, after filling it with water, and thus a real incendiary lentil was obtained, with the help of which, concentrating the rays on dry moss, the engineer produced fire.”

What work is this excerpt from?

What physical device did the engineer make?

Why was the engineer able to light the fire?

4. Guys, on the surface of this lake (showing a photograph of a sunset near a pond ) a sparkling path is visible against the sun. How is it formed? Why is the track always oriented towards the observer?

(The path on the surface of the water occurs due to the reflection of light from small waves that are oriented in various directions. Therefore, at the most various positions The observer's reflected rays enter his eye. Each observer sees “his” path.)

5. Explain from the point of view of physics the observed phenomenon described by Bunin in the quatrain.

And at noon there are puddles under the window
So they spill and shine,
What a bright sunspot
The “bunnies” are fluttering around the hall.
I.A. Bunin.

IV. Lesson summary:

Today in class we:

repeated the basic concepts on the topic "Optics"

remembered

law of reflection

law of light refraction

constructing an image in lenses and mirrors

explained some natural phenomena

applied our knowledge of “Optics” when answering questions

There will be a test on this topic in the next lesson. I wish you only good and excellent grades.

You have to study a lot to know even a little.
Charles Louis Montesquieu
Montesquieu Charles Louis de Secondat(01/18/1689– 02/10/1755) - French educator, jurist, philosopher and writer.

BOX OF QUALITY PROBLEMS IN PHYSICS
OPTICS (LIGHT PHENOMENA)

Didactic materials on physics for students, as well as their parents;-) and, of course, for creative teachers. For those who love to learn!

The task box has been updated as a result exciting work above the green pages of the section "Physics and fiction: Optics"

Problem No. 46
A young fisherman, sitting on the shore of a lake, sees an image on the smooth surface of the water morning sun. Where will this image move if he watches it while standing?

Answer: Move away from the shore.

Problem No. 47
A scuba diver swimming underwater can always see a fisherman on the shore. A fisherman sitting on the shore can only rarely see a scuba diver swimming under water. Why?

Answer: Light reflected from a scuba diver large angles falling rays experiences complete reflection from the water-air boundary. The light reflected from the fisherman at any angle of incidence passes into the water.

Walter Dandy Sadler(Walter Dendy Sadler; 05/12/1854–11/13/1923) - an English painter who specialized in everyday and characteristic, often comic scenes from the 18th century.

Problem No. 48
If you stand on a mountain with your back to the sun and look into the thick fog spreading in front of you, you can see a rainbow border (or a closed ring) around the shadow of the head. Why does a halo appear and how are the colors arranged in it?

Answer: The halo occurs due to the reverse (towards the source) scattering of light by water droplets, the dimensions of which are commensurate with the wavelength of the light. The returning light enters the droplet from the side and from the side, but on the other hand it exits, having undergone reflection inside the droplet, as well as going around it along the surface (diffraction). The backscattering angle depends on the wavelength, so colored rings are formed; Since the angle also depends on the size of the droplets, rings appear only when the droplets do not differ much in size.

Problem No. 49
"Typhoon off the coast of Japan", 1893. Jack London
“The morning was magnificent, but our helmsman, looking at rising sun, shook his head cautiously and muttered meaningfully: “The red sun in the morning is not to the sailor’s liking.” Indeed, the sun looked so ominous that several light curly clouds frolicking in the sky, as if frightened by it, hastily disappeared somewhere ... "
Explain the words of the helmsman: “The sun is red in the morning for a sailor”? Why does the Sun play different colors at sunrise and especially sunset?

Answer: It must be said that there are many proverbs in which red morning sky seen as a warning of rain. Shakespeare wrote that a red morning always foretells a storm to the fields, and a shipwreck to the sailor. The red color of the sun portends strong winds, especially in the upper layers of the atmosphere and is determined by the presence of dust and humidity accompanying rain in the atmosphere.
The sun's rays travel a long way in the air at sunset and sunrise. According to Rayleigh's theory, blue, cyan and violet rays will be scattered, and rays of the red part of the spectrum will pass through. Therefore, the Sun is painted in yellow, pink, red tones, opposite side the sky seems to be colored blue with purple tint color. Sunrise gives a brighter and cleaner picture, as the air becomes cleaner overnight.

Problem #50
If we look at the surrounding bodies through the warm air rising from the fire, then they seem to us to be trembling. Why?

Answer: The refractive index of air above a burning fire changes depending on the air temperature. The air heats up unevenly and becomes inhomogeneous. Air layers are constantly moving. The air above the burning fire heats up, expands and rises, and colder air rushes in its place. Rays of light in such a heterogeneous environment are refracted differently, the picture changes all the time - the air rising from the fire trembles and flows, causing the surrounding bodies to tremble and distort...

Problem No. 51
What source of light for us is twilight, morning and evening dawns?

Answer: Sources of scattered sunlight, molecules of atmospheric gases and dust particles in the atmosphere.

Zankovsky Ilya Nikolaevich(1832–1919) - Russian landscape artist, master of majestic mountain landscapes.

Rose gold is an alloy of gold, copper and silver, in which silver contains approximately five times less than gold, and the amount of copper is relatively large.

§ Small gallery paintings by Ilya Nikolaevich Zankovsky on the green page “Photo Album: Mountain Landscapes (Caucasus)”
Tourist tour: “Spring in the mountains”, May 2008

Krause Franz Emil(Krause Franz Emil; 1836–1900) – German painter, landscape painter.

Red gold is a high-grade alloy of gold and copper, which was previously used to make chervonets and other coins. Red gold is also sometimes called gold. of the highest standard bright yellow, the so-called pure gold.

Problem No. 52
You and I can look at the sun when it is near the horizon, but we cannot when it is high. Why?

Answer: The intensity of the sun's rays at sunset or sunrise is much less than during the day, since the rays pass through a thicker layer of air at this time and are more absorbed.

Problem No. 53
How to explain the appearance of a rainbow after rain? Why does a rainbow have an arc shape?

Answer: The rainbow can be considered as a giant “colored wheel”, which, like an axle, is “put on” an imaginary straight line passing through the sun and the observer. In all its glory, this phenomenon can only be observed from an airplane. In this case, for an observer, the rainbow will look like a rainbow circle with the shadow of an airplane in the center.

Problem No. 54
When the rainbow is higher:
at 4 or 5 o'clock in the afternoon?

Answer: At 5 o'clock, since the lower the Sun goes, the higher point O rises (Fig. 1). Consequently, the rainbow rises above the horizon in an increasingly larger arc, and when the Sun sets, it becomes semicircular.

Problem No. 55
Is it possible to see a rainbow if you are at one end?

Answer: It is forbidden. The observer's eye is always in a plane passing through the center of the rainbow and the center of the solar disk (Fig. 1).

Dubovskoy Nikolay Nikanorovich(12/17/1859–02/28/1918) - Russian painter, member of the Association of the Wanderers.

Problem No. 56
How far from us does the rainbow form, i.e. At what distance are those drops of water due to which it arises?

Answer: For a rainbow, all that matters is the angle between the incident sunbeam and the observer's line of sight. Drops can be located at a distance from several meters to several kilometers. You can observe a rainbow appearing against the background of a waterfall or fountain.

Frederic Edwin Church(Frederic Edwin Church: 05/04/1826–04/07/1900) - American landscape artist.

Problem No. 57
Why do dew droplets glisten? Why do air bubbles shine in water?

Answer: Due to the phenomenon of total reflection.

Problem No. 58
A traveler who climbs a steppe hillock on a quiet sunny morning sometimes thinks that distant objects are reflected in the water surface (mirage). This illusion occurs because rays coming from objects in the direction of the soil experience complete reflection from the ground layer of air and only then enter the eye of the observer. What effect does heating the air have on the refractive index?

Answer: Total reflection can only occur from an optically less dense medium. The sun's rays heat the lower layers of the atmosphere. Their refractive index decreases.

Problem No. 59
Why are ordinary clouds mostly white and storm clouds black?

Answer: The size of water droplets in a cloud is much more molecules air, so the light from them is not scattered, but reflected. However, it does not decompose into its components, but remains white. Very dense thunderclouds either do not transmit light at all or reflect it upward.

Kryzhitsky Konstantin Yakovlevich(1858–1911) – Russian landscape painter, academician Imperial Academy Arts

Savrasov Alexey Kondratievich(05/12/1830–09/26/1897) - Russian landscape painter, one of the founding members of the Partnership of the Wanderers.

For the curious:
The formation of clouds is impossible without particles of dust or smoke, around which moisture begins to condense.
The mass of 1 km 2 clouds is ≈ 2,000,000 kg.
Height of large thunder clouds can reach 10 km.
The highest of the clouds are noctilucent clouds. They arise in the upper layers of the atmosphere, at an altitude of 70-80 kilometers. You can see them only at dusk. A charming and exciting spectacle!

Problem No. 60
Sometimes pearlescent clouds appear that have very beautiful tones. After sunset they are so bright that the light from them colors the snow. What are the features of these clouds?

Answer: Pearlescent clouds are observed mainly in polar latitudes at altitudes from 15 to 27 km and are a fairly rare phenomenon. Mother-of-pearl clouds consist of droplets whose radii (0.1–3 µm) are close to the wavelength visible light. Light diffraction occurs on these drops, which depends on the radius of the drop and the wavelength.

Problem No. 61
Why doesn't the whole sky have the same hue, but part of it is painted a brighter blue?

Answer: Sunlight is scattered by air molecules, with shorter wavelengths being scattered more strongly. Therefore, when the Sun is close to the horizon, the sky above the observer is mostly blue. The blueness of the sky at a distance of more than 90° from the Sun is weaker, since the sky is illuminated by light that has traveled a greater distance in the atmosphere and has lost its blue component.

§ Read interesting things about the color of the sky at. Translation of “Rayleigh scattering of sunlight” into Russian;-)

Problem No. 62
The sun's rays breaking through the clouds appear radially diverging in all directions. Meanwhile, the sun's rays falling on the earth are parallel. How to explain such a contradiction?

Answer: The divergence of the rays is explained by the perspective effect of the convergence of receding parallel lines.

Gine Alexander Vasilievich(1830–1880) - Russian painter and graphic artist, landscape painter, illustrator, friend and fellow student of Ivan Ivanovich Shishkin.

Problem No. 63
Why does darkness set in faster when the sky is cloudless than when there are clouds?

Answer: For some time after the Sun sets below the horizon, the surface of the Earth is illuminated by light reflected from the clouds.

Problem No. 64
What's bigger: the cloud or its complete shadow?

Answer: The cloud casts a cone of complete shadow tapering towards the ground, but the height of the cone due to significant size the clouds are very large. Therefore, the total shadow of a cloud on Earth differs practically little in size from the cloud itself.

Problem No. 65
Is it possible to get four penumbra without a shadow from an opaque object?

Answer: Can. Imagine a football player in the center of the stadium, illuminated in the evening by floodlights mounted high in the corners of the stadium.

Problem No. 66
Can the length of the shadow from a rope stretched between vertical posts be greater distance between the pillars?

Answer: Yes, it can if the rope is stretched, for example, over a ravine.

Problem No. 67
“Who Lives Well in Rus',” 1865–1877
Nikolai Alekseevich Nekrasov
“...The moon has surfaced, the shadows are black
The road was cut
Zealous walkers.
Oh shadows! black shadows!
Who won't you catch up with?
Who won't you overtake?
Only you, black shadows,
You can’t catch or hug!”
Can a person outrun his shadow?

Answer: Maybe if a shadow forms on a wall, parallel to which a person is moving, and the light source is moving faster than the person and in the same direction.

Levitan Isaac Ilyich(08/30/1860–07/22/1900) - Russian landscape painter - great and consummate master“landscape of mood”, member of the Association of Itinerants.

Problem No. 68
Road irregularities are less visible during the day than at night when the road is illuminated by car headlights. Why?

Answer: When the road is illuminated with headlights, road unevenness creates shadows that are clearly visible from a distance.

Problem No. 69
In some rooms, lighting fixtures are positioned so that the light they create does not fall directly on the workplace, but illuminates the white ceiling of the room. What are the advantages of this lighting method?

Answer: With this lighting, sharp shadows are not produced.

Problem No. 70
What explains high value diamond as a material for jewelry and why it cannot be used complete success replace it with glass in this case?

Answer: The refractive index of diamond is significantly higher than that of glass; Most of the rays entering a diamond experience complete internal reflection in it.

Problem No. 71
There is an opinion that diamonds become invisible in water. Is this true?

Answer: This is a misconception. An object can become invisible if it is surrounded by a substance that has the same refractive index as itself. The refractive index of diamond is 2.42, while the refractive index of water is 1.33. Even ordinary glass with a refractive index of 1.5 is also noticeable in water.

Problem No. 72
In H.G. Wells's novel The Invisible Man, the main character invented a special compound and, after drinking it, became completely transparent to light rays, and therefore invisible. In the novel, the invisible man himself sees everything around him, while remaining invisible. Can such a person see?

Answer: In order for a person to see, light must be absorbed at the retina of the eye. But if a person has become completely transparent, then his retina is transparent, which means it does not absorb light. In addition, in order for a person to see, and not just feel light, an image of objects must be obtained on the retina of the eye. Therefore, if the membranes of the eye become transparent, then the person will also lose the ability to see images, because light will hit the retina in addition to the pupil.

“...Then he took off his glasses, and everyone’s eyes widened in surprise. He took off his hat and began to furiously tear off his sideburns and bandages. They did not immediately succumb to his efforts. Everyone froze in horror.
- Oh my God! - someone said.
Finally the bandages were torn off.
What appeared before the eyes of those present exceeded all expectations. Mrs. Hall, who stood with her mouth open, screamed wildly and ran towards the door. Everyone jumped up from their seats. They expected wounds, ugliness, visible horror, but here - nothing. The bandages and wig flew into the tavern, almost hitting those standing there. Everyone rushed away from the porch, bumping into each other, because on the threshold of the living room, shouting incoherent explanations, stood a figure that looked like a man right down to the collar of his coat, and above there was nothing. Absolutely nothing!”


Translation: D. Weiss

“...The Invisible Man asked for a cigar. He bit off the tip greedily before Kemp had time to find the knife, and cursed when a leaf of tobacco fell away from the outside. It was strange to see how he smoked: his mouth, throat, pharynx and nostrils appeared, like a cast made of billowing smoke...”
"The Invisible Man", 1897, H. G. Wells
The Invisible Man (1897) Herbert Wells
Translation: D. Weiss

Herbert George Wells(Herbert George Wells; 09/21/1866–08/13/1946) – English writer and publicist. Author of famous science fiction novels “The Time Machine”, “The Invisible Man”, “War of the Worlds”...
Itkin Anatoly Zinovievich(1931...) - Soviet, Russian graphic artist, illustrator, working in the historical adventure genre, Honored Artist of Russia.

I strongly recommend that readers of the green pages read or re-read H.G. Wells' novel The Invisible Man. Modern film adaptations of the novel excessively abuse special effects, some of which are not at all friendly with physics:-(They strangely distort storyline, pushing the ideas of H.G. Wells himself to the very rear:-(

Problem No. 73
There are organisms (larvae of some insects, for example, the larva of the pinnate mosquito) that are not visible in water because of their transparency. But the eyes of such invisible creatures are clearly visible in the form of black dots. Why are these creatures not visible in the water? Why are their eyes opaque? Will they remain invisible in the air?

Answer: The refractive index of the insect's body is close to that of water, but the refractive index of the eyes is different. Light would pass through transparent eyes without being refracted, and images would not be formed on the retina. Larvae are visible in the air.

Problem No. 74
If you look at multi-colored luminous advertising (for example, from gas-discharge tubes), the red letters always seem to protrude in comparison with the blue and green ones. How can this be explained?

Answer: The focal length of the eye, like any lens, is different for different wavelengths, that is, for different colors spectrum Red rays are refracted less strongly, so the visual impression arises that red objects are closer to the observer than blue ones.

Problem No. 75
At low light blue appears brighter than red, but in good lighting, red appears brighter than blue. Why does the relative brightness of colors depend on the level of light?

Answer: In strong light conditions, vision is determined by cones, and in low light conditions, by rods. There are three types of cones that are sensitive to colors: red, yellow, and blue. Rods are most sensitive to green light and less sensitive to red. If you increase the illumination, your vision switches from “rod” to “cone” vision. Red colors at dusk appear darker than green ones, and at night they appear almost black, while blue objects “become” lighter (Purkinje effect).

Problem No. 76
Which is lighter: black velvet on a sunny day or pure snow on a moonlit night?

Answer: Black velvet in sunlight is many times lighter than snow illuminated by the Moon.

§ A detailed answer to this problem performed by Yakov Isidorovich Perelman can be found at

Remigius van Haanen(Remigius Adrianus van Haanen; 01/05/1812–08/13/1894) - Dutch painter.

Problem No. 77
Why does dirty snow melt faster than clean snow?

Answer: The ray absorption coefficient of dirty snow is greater than that of clean snow.

Problem No. 78
Why is the color red used for prohibited signals on transport?

Answer: Red rays travel with less loss. Therefore, the red signal is visible further.

Problem No. 79
Illuminated from a certain direction, a single-color flag flying on a building at night appears striped, and the stripes on it are constantly moving. What are the reasons for this phenomenon?

Answer: The brightness of the illuminated area of ​​​​the fabric depends on its illumination, and the illumination depends on the cosine of the angle of incidence of the rays; in the case of one-sided illumination, this angle is different for different parts of the waving flag.

Problem No. 80
Which of two decorative lamps of the same power - red or green - emits a greater luminous flux?

Answer: Green. The human eye is most sensitive to green color, and the luminous flux is assessed precisely by visual sensation.

Marcel Rieder(Marcel Rieder; 03/19/1862–03/30/1942) - French painter.

Problem No. 81
What function does a lampshade serve for a table lamp or floor lamp?

Answer: The lampshade is designed to protect the eyes from the glare of a light source and create the required illumination by reflecting/absorbing/scattering it.

For the curious:
Shade translated from French abat-jour - light muffler. The lampshade first appeared in France more than two hundred years ago and since then it has confidently and for a long time conquered the whole world, firmly establishing itself in the homes of many people. Since ancient times, to protect the eyes from the blinding light of torches, candles, kerosene lamps, and later incandescent lamps... people invented special dampers that absorb and scatter light. Gradually, they began to pay attention to this decorative element special attention professional designers and decorators. Lampshades began to be made in the most unimaginable shapes and shades; from different types fabrics, glass, plastic... decorate with frills, fringe, beads and other interesting things... Thus, the lampshade has become not just a necessary and functional part of the interior, but has also reached a new, very impressive level ;-) giving people inimitable aesthetic pleasure.

Problem No. 82
Why do the windows of houses appear dark during the day, that is, darker than the outer walls, even if the walls are painted dark?

Answer: Because the reflection of light from the walls is always greater than the reflection from transparent, that is, windows that transmit light.

Fernand Toussaint(Fernand Toussaint; 1873–1956) – Belgian painter.

Problem No. 83
Why is dry sand light, but wet sand appears dark?

Answer: Wet sand appears dark because the reflection from the sand is greatly reduced and most of the light passes inward where it is absorbed.

Problem No. 84
Why does the forest visible on the horizon seem not green, but shrouded in a bluish haze?

Answer: Blue and blue rays are scattered more strongly than others by air. Therefore, the layer of air between the observer and the distant forest appears bluish, like the sky.

Problem No. 85
The rich blue color of copper sulfate crystals becomes light turquoise if these crystals are ground into a fine powder; red-orange crystals of potassium dichromate under the same conditions give a pale yellow powder. How to explain these phenomena?

Answer: Finely crushed transparent substance strongly scatters the light falling on it; Thus, light in the crushed substance does not penetrate to great depths, and therefore is little absorbed.

Problem No. 86
What will a copper sulfate solution look like when illuminated with red light? green? purple?

Answer: Purple (almost black), green, blue.

Problem No. 87
At your disposal are transparent crimson, blue and yellow glasses. What colors can be obtained by combining these glasses?

Answer: Crimson and yellow let in red rays; blue and yellow – green; blue with crimson - purple; all three together give the color black.

Herbert James Draper(Herbert James Draper; 1863–1920) – English painter.

For the curious:
Stained glass(French vitrage, from Latin vitrum glass), an ornamental or narrative decorative composition (in a window, door, partition, in the form of an independent panel) made of glass or other material that transmits light. The birth of stained glass as such can be dated approximately to the 6th–7th centuries. AD, when the Christian Catholic Church began to use stained glass to create a special emotional atmosphere.
Classic stained glass windows are made from glass, to which coloring metal oxides are added during its production. The resulting sheets of colored glass are cut into small pieces of the required shape, from which the intended pattern or picture is then laid out. These pieces are connected and held together by flexible strips of lead that form dark, expressive outlines.
The main advantage of stained glass is that it is exposed to light, maintaining maximum color intensity and is capable of delightfully transforming and enlivening the surrounding space.
The oldest surviving example of stained glass is in the cathedral of Augsburg, Germany, and dates back to approximately 1100.

§ I invite readers to look at the green page “Photo Album: Mom’s Garden – Flower Kaleidoscope”. Here are flower photographs for your attention, accompanied by educational interests and a little physics;-) What it will look like blooming garden, if viewed through colored glass? What color do red gladioli appear through green glass? And blue irises and green leaves - through the same glass?

Problem No. 88
A candle flame viewed through steam appears red to us. Why?

Answer: Because steam scatters rays that have a shorter wavelength (violet, blue, cyan, green, yellow).

Problem No. 89
Mikhailo Vasilyevich Lomonosov in one of his notes poses the following question: “When wetted with water, any color becomes thicker. Why? We need to think." How to answer this question?

Answer: Surface color is determined spectral composition rays reflected by it. When the surface is dry, scattered light is added to the rays corresponding to the color of the surface. white light from surface unevenness. Therefore, the surface color appears less bright. When the surface is saturated with water, the irregularities are covered by a surface film of water and the scattered radiation disappears. Therefore, the basic tone of the surface color is perceived by us as darker.

Problem No. 90
Why do colored fabrics fade in the sun?

Answer: Ultraviolet radiation, being absorbed organic molecules paints, disrupts molecular bonds. This leads to loss of pigment.

Jean-Baptiste Jules Trayer(Jean-Baptiste Jules Trayer; 08/20/1824–01/01/1909) - French painter.

Problem No. 91
Flame electric arc It will be harmless to vision if the arc is lit in water. Why?

Answer: Water absorbs ultraviolet rays.

Problem No. 92
For protection from the sun's rays, white and red umbrellas are the most practical. Why?

Answer: These umbrellas reflect orange, red and infrared rays well.

Karl Schwenninger the Younger(Carl Schweninger der Jüngere; 05/17/1854–12/27/1912) - Austrian painter.

Problem No. 93
The faces of climbers at high altitudes become very tanned in a short time. Why?

Answer: Air strongly scatters ultraviolet rays. At high altitudes, where the air is thin, ultraviolet radiation is very intense.

Problem No. 94
The spotlight's beam is clearly visible in fog, but worse in clear weather. Why?
Why do the beams of searchlights aimed at the night sky (for example, those used during the war to detect aircraft) end so abruptly in the air?

Answer: The spotlight beam is clearly visible in fog due to the scattering (reflection) of light by small droplets of water.
The searchlight beam weakens not only due to divergence, but also due to atmospheric scattering. Therefore, its intensity decreases exponentially and ends quite abruptly.

Problem No. 95
How can we explain the formation of colored spots on the surface of the water in places where it is contaminated with oil, gasoline or lubricating oil?

Answer: Rainbow stripes in thin films arise from the interference of light waves reflected from the upper and lower boundaries of the film. The wave reflected from the lower boundary lags in phase from the wave reflected from upper limit. The magnitude of this lag depends on the thickness of the film and on the length of the light waves in the film. Due to interference, some colors of the spectrum will be damped and others will be enhanced. Therefore, areas of the film with different thicknesses will be painted in different colors.

Problem No. 96
On the glass, long time exposed atmospheric influences or lying for a long time in damp soil, beautiful rainbow shades are observed. How to explain their origin?

Answer: On the surface of the glass, as a result of corrosion (rainbow leaching), thin layer glass of a different composition - “colors” of thin films appear.

For the curious:
Pearlescent and iridescent spots on the glass surface can appear not only as a result of improper storage and, as a result, glass corrosion (rainbow leaching), but can also be caused by certain features of glass manufacturing technology. As a result of violations technological process, the surface of the glass may become contaminated, for example, with mineral oil that was not removed during washing. Due to the phenomenon of interference in thin films on glass, rainbow stains can be observed in this case irregular shape, similar to those that can often be seen on the surface of water contaminated with oil or gasoline.

Problem No. 97
Under the influence of heating to a temperature of 220-350°C, steel is covered with a brightly colored multi-colored film, the so-called “tarnish colors”. Explain the phenomenon.

Answer: At temperatures of 220-350°C, steel is coated with a thin transparent layer oxide. The thickness of this layer (and therefore the color of the tarnish) depends on the temperature. For example, a temperature of 220°C corresponds to a light yellow color, a temperature of 285°C – purple.

Problem No. 98
The shell of a soap bubble is in some places straw-yellow in color, in others crimson, in others greenish-blue. Why do these differences in color occur and what is the approximate thickness of the film that forms the membrane of the bubble?

Answer: The thickness of the film in areas with a yellow color is about 0.15 microns; in blue – almost twice as much. The same colors can also be observed in places where the film thickness is expressed in multiples of these values.

George Sheridan Knowles(George Sheridan Knowles; 11/25/1863–03/15/1931) - English painter.

Problem No. 99
When observing soap film formed in a flat vertical frame, you can notice that the colored horizontal interference fringes will move downward over time, slightly changing their width. After some time, a rapidly increasing black spot, and then the film will tear. Explain what you observed.

Answer: Water in inner layer The film gradually flows down, the lower part of the film thickens, and the upper becomes thinner. Places corresponding to a certain film thickness move, and the corresponding interference fringes move along with them. After some time, the thickness of the film in the upper part becomes less than a quarter of the wavelength of the most short waves light falling on the film. In these places of the film, due to the interference of rays reflected from the film, waves of all lengths will be damped.

Problem #100
Small spots appearing on the leaves of plants are sometimes observed. sunny days in those places where drops of water remained after rain or watering. What is the reason for the appearance of such spots?

Answer: A drop of water is a small lens, which, collecting the sun's rays into focus, causes a slight burn on the surface of the leaf. For this reason, experienced gardeners and gardeners water their beds in the morning or evening.

Edmond Louis(Edmond Louyot; 11/15/1861–01/17/1920) - French painter.

Problem No. 101
If you light a candle in front of a dusty mirror, you can see a rainbow halo around the flame. Carry out an experiment and explain the observed phenomenon.

Answer: The appearance of a rainbow halo around a candle flame is due to the phenomenon of diffraction.

Problem No. 102
When making artificial mother-of-pearl buttons, the finest shading is applied to their surface. Why does the button have a rainbow color after this treatment?

Answer: The smallest shading plays the role of a diffraction grating, producing a spectrum in reflected rays.

Problem No. 103
Steelworkers have to work in difficult conditions when they deal with molten metal: its hot breath literally burns a person. It would seem that to facilitate working conditions, the suits of blast furnace workers, open-hearth furnaces and other metallurgists should be made of materials with low thermal conductivity. Meanwhile, in fact, modern workwear for metallurgists is often covered with a layer of metal - an excellent heat conductor. For what purpose are they doing this?

Answer: The transfer of heat from hot metal to a person occurs mainly through radiation. The maximum radiation energy at metal temperature is carried by infrared rays, which, as in general electromagnetic waves, are very strongly reflected by metals. This answers the question why steelworkers’ clothes are metalized.

Peder Severin Kroyer(Peder Severin Krøyer; 07/23/1851–11/21/1909) - Danish painter.

Problem No. 104
Why, when looking at a row of lamps located along a long street, do we see them equally bright, although the distance from the eyes to the lamps is not the same?

Answer: The apparent brightness of the flashlight is equal to the illumination of the image on the retina, that is, the ratio of the light flux entering the eye to the area of ​​the image on the retina. As the distance to the light source increases, the light flux entering the eye decreases, but at the same time the image area on the retina also decreases. The ratio of these two quantities remains constant if the loss of light energy due to absorption and scattering of light as it propagates through the air can be neglected. In fog, the apparent brightness of the image decreases as the light source moves away as energy absorption and dissipation become noticeable.

Problem No. 105
On a light background ceramic product a dark drawing was made. If this product is placed in an oven with high temperature, then a light pattern is visible on a dark background. Why?

Answer: Since the pattern is dark, it radiates more strongly than a light ceramic product.

Final chord ;-) Problem – Picture – Riddle

MIRROR OR PICTURE?

Stefan Sedlacek(Stephan Sedlacek, 1868-1936) - Austrian painter.

In Stefan Sedlacek's painting we see not just a painting within a painting, but maybe a huge mirror??? The magnificent crystal chandelier, the chair with bent animal legs on the right in the direction are confusing... and, in fact, the angle itself ;-)
What can you say on this topic? Write your versions in the comments ;-)

I wish you success in making your own decision
quality problems in physics!

Literature:
§ Collection of problems in physics for grades VIII–X of secondary school
edited by Znamensky P.A.
Moscow: publishing house "UCHPEDGIZ", 1951
§ Tulchinsky M.E. Qualitative problems in physics
Moscow: Prosveshchenie Publishing House, 1972
§ Zolotov V.A. Questions and tasks in physics for eight-year school
Moscow: Prosveshchenie Publishing House, 1965
§ Demkovich V.P., Demkovich L.P. Collection of problems in physics
Moscow: Prosveshchenie Publishing House, 1981
§ Shaskolskaya M.P., Eltsin I.A. Collection of selected problems in physics
Moscow: Nauka Publishing House, 1967
§ Lukashik V.I. Physics Olympiad
Moscow: Prosveshchenie Publishing House, 1987
§ Katz Ts.B. Biophysics in physics lessons

§ Perelman Ya.I. Do you know physics?
Domodedovo: publishing house "VAP", 1994
§ Lange V.N. Physical paradoxes and sophisms
Moscow: Prosveshchenie Publishing House, 1978
§ Tarasov L.V. Physics in nature
Moscow: Prosveshchenie Publishing House, 1988

Option 1

1. What law is the formation of a shadow evidenced by?

A. Only the law of light refraction.

B. Only the law of light reflection.

B. Only the law rectilinear law propagation of light.

D. All three laws.

2. It is known that we see bodies that are not sources of light. What phenomenon leads to this?

A. Reflection of light.

B. Refraction of light.

B. Light absorption.

D. All three phenomena

3. Determine the optical power of a converging lens whose focal length is 40 cm.

A. 0.25 diopters. B. 2.5 diopters. V. 0.4 diopters. G. 0.05 diopters.

4. What image is obtained on the retina of the eye?

A. Real, imaginary. B. Imaginary, direct. B. Real, inverted. G. Imaginary, inverted.

5. At what distance is the subject usually located when photographing in relation to the lens with a focal lengthF?

A. l > FB. F< l < 2F IN. l > FG. l < F

6. How will the angle between the incident and reflected rays of light change if the angle of incidence decreases by 10°?

A. Will decrease by 5°.

B. Will decrease by 10°.

B. Will decrease by 20°.

G. It won’t change.

7. A person stands in front of a vertically placed flat mirror at a distance of 1 m from its plane. What is the distance between the image of a person and the mirror?

A. 0.5 m. B. 1 m. C. 2 m. D. 4 m.

8. A person is standing in front of a vertically placed flat mirror. How will the distance between a person and his image change if the person approaches the plane of the mirror by 1 m?

A. Will decrease by 2 m. B. Will decrease by 1 m. C. Will decrease by 0.5 m. D. Will not change.

9. What is the focus of a converging lens?

A. The point at which the rays refracted by the lens intersect.

B. The place where the lens collects all the rays.

B. The point on the optical axis of the lens at which the rays of light refracted by it and incident on the lens parallel to the optical axis intersect.

D. The point at which the continuations of the rays are collected.

10. The focal length of one lens is shorter than the other. Which one has more surface curvature?

A. Short-focus.

B. Identical.

B. At the long focal length.

D. The curvature of the surface does not depend on the focal length.

Option 2.

1. What should the light source be like so that there is shadow and penumbra behind the object illuminated by it?

A. Point. B. Extended. B. Anyone. G. Yarkim

2. It is known that bodies that are not sources of light can be visible. What phenomenon leads to this?

A. Light absorption.

B. Reflection of light.

B. Refraction of light.

D. None of the phenomena

3. Determine the optical power of a converging lens whose focal length is 50 cm.

A. 0.2 diopters. B. 2 diopters. B. 20 diopters. G. 0.02 diopters.

4. What image of an object does a flat mirror give?

A. Imaginary, behind the mirror, at the same distance from it as the object, and the same size as it.

B. Real, in front of the mirror further than the object, and smaller in size than it.

B. Imaginary, at different distances from it behind the mirror and different sizes depending on where the object is in front of the mirror.

D. Real, behind the mirror, at the same distance from it as the object, and the same size as it.

5. The optical power of some glasses (1) is - 2 diopters, others (2) + 2 diopters. What eyes are these glasses intended for?

A. 1- for myopic, 2- for farsighted.

B. 1- for farsighted, 2- for nearsighted.

B. Both 1 and 2 are for myopic people.

G. Both 1 and 2 are for farsighted people.

6. How will the angle between the incident and reflected rays of light change if the angle of incidence is increased by 20°?

A. Increase by 40°.

B. Increase by 20°.

B. Increase by 10°.

G. It won’t change.

7. A person stands in front of a vertically placed flat mirror at a distance of 2 m from its plane. What is the distance between the image of a person and the mirror?

A. 8 m. B. 4 m. C. 2 m. D. 1 m.

8. A person is standing in front of a vertically placed flat mirror. How will the distance between a person and his image change if the person moves 2 m away from the plane of the mirror?

A. It won't change. B. Increase by 1 m. C. Increase by 2 m. D. Increase by 4 m.

9. Which lens - concave or convex - is a converging lens?

A. Concave. B. Convex. B. Concave-convex. D. All types of lenses collect light.

10. At what distancedof an object from a converging lens, will its image be real, inverted and enlarged?

A. Whend < F

B. WhenF< d < 2 F

B. Whend > 2 F

G. Whend = F

Answers

This benefit fully complies with the federal state educational standard(second generation).
The manual is intended to test students' knowledge of the 8th grade physics course. It is focused on the textbook by A.V. Peryshkin “Physics. 8th grade" and contains tests on all topics studied in 8th grade, as well as independent work to each paragraph.
Tests are given in four versions, each option includes tasks of three levels, which corresponds to the forms of tasks used in the Unified State Exam. The manual will help quickly identify gaps in knowledge and is addressed to both physics teachers and students for self-control.

Examples.
50 liters of water at a temperature of 15°C and 30 liters of water at a temperature of 75°C were poured into the bath and mixed. Determine the established temperature. Neglect energy losses. The density of water is 1000 kg/m.

To bathe a child, 40 liters were poured into the bath cold water, the temperature of which was 6 °C, and then added hot water at a temperature of 96 °C. Determine the mass of hot water if the temperature of the water in the bathroom becomes 36 °C. The density of water is 1000 kg/m.

200 g of boiling water was poured into a porcelain cup weighing 100 g at a temperature of 20 °C. The final temperature was 93 °C. Determine the specific heat capacity of porcelain. Specific heat water 4200 J/(kg °C).

TABLE OF CONTENTS
Chapter 1-2. Thermal phenomena. Change of aggregative states of matter 5
INDEPENDENT WORK 5
SR-1. Thermal movement. Temperature. Internal energy 5
SR-2. Ways to change internal energy body 7
SR-3. Thermal conductivity 8
SR-4. Convection 9
SR-5. Radiation 10
SR-6. Amount of heat. Specific heat. Calculation of the amount of heat required to heat a body or released by it during cooling 11
SR- 7. Heat exchange (without aggregate transitions) 12
SR-8. Fuel energy. Specific heat combustion 13
SR-9. Law of conservation and transformation of energy in mechanical and thermal processes 14
SR-10. Melting and solidification crystalline bodies 15
SR-11. Graph of melting and solidification of crystalline solids 16
SR-12. Specific heat of fusion 17
SR-13. Evaporation. Rich and unsaturated steam 18
SR-14. Energy absorption during liquid evaporation and its release during steam condensation 19
SR-15. Boiling 20
SR-16. Air humidity. Methods for determining air humidity 21
SR-17. Specific heat of vaporization and condensation 22
SR-18. Thermal processes 23
SR-19. Heat transfer (with aggregate transitions) 25
SR-20. Heat engine efficiency 27
CHECK WORK 28
Option No. 1 28
Option No. 2 31
Option No. 3 34
Option No. 4 37
Chapter 3. Electrical phenomena 40
INDEPENDENT WORK 40
SR-21. Electrification of bodies upon contact 40
SR-22. Interaction of charged bodies. Two types of charges. Electroscope. Conductors and non-conductors of electricity. Electric field 42
SR-23. Atomic structure 43
SR-24. Explanation electrical phenomena 44
CONTROL WORK “ELECTRICAL PHENOMENA” 45
Option No. 1 45
Option No. 2 48
Option No. 3 51
Option No. 4 54
Chapter 3 (continued). DC 57
INDEPENDENT WORK 57
SR-25. Electric current. Sources electric current 57
SR-26. Actions of electric current 58
SR-27. Current strength. Current units 59
SR-28. Electrical voltage. Voltage units 60
SR-29. Electrical resistance conductors. Units of resistance. Calculation of conductor resistance. Resistivity 61
SR-30. Ohm's law for circuit section 62
SR-31. Calculation of impedance and current in a circuit 63
SR-32. Calculation of electrical circuits 65
SR-33. Work and power of electric current 68
SR-34. Heating of conductors by electric current.
Joule-Lenz Law 69
CONTROL WORK “DC CURRENT” 70
Option No. 1 70
Option No. 2 73
Option No. 3 75
Option No. 4 77
Chapter 4. Electromagnetic phenomena 79
INDEPENDENT WORK 79
SR-35. Electromagnetic phenomena 79
CHECK WORK 80
Option No. 1 80
Option No. 2 83
Option No. 3 86
Option No. 4 89
Chapter 5. Light phenomena 92
INDEPENDENT WORK 92
SR-36. Light sources 92
SR-37. Propagation of Light 93
SR-38. Reflection of light. Laws of reflection. Flat mirror 94
SR-39. Refraction of light 95
SR-40. Lenses. Optical power of the lens. Images produced by lens 96
CHECK WORK 98
Option No. 1 98
Option No. 2 100
Option No. 3 102
Option No. 4 104
ANSWERS 107.

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