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22 -- Part 1: Mirrors and Lenses
Two of your roommates wear glasses.
Komila needs them to drive and Jason needs them to read and write.
The ones labeled as + 1.5 are the ones with -2.0.
The properties of similar triangles can be applied.
You will learn how to answer the questions in this chapter.
The laws of reflection and refraction can be used.
We will learn how cameras, telescopes, microscopes, human vision, and corrective lens work.
Light coming from an object.
You can see a reflection of yourself when you stand in front of a plane mirror.
We created a model to describe how objects emit light.
Imagine that the shining object is in front of a light source and the origination point is a mirror.
There is a point object in the plane mirror.
They see the rays in the mirror.
At one point behind the mirror, the dashed extensions of the solid lines intersect.
The image behind the mirror is not on the mirror's surface.
The image behind the plane mirror is exactly the same distance as the object in front of it.
The image location of a plane mirror is being tested.
Some of the reflected rays will still reach where they were before, but we can see the image of the bulb in the experiment part of the mirror.
The location of the image should not change.
The mirror with the clean sheet behind it is exactly the same distance behind the bulb as it is in the glass.
Bulb 2 should be behind the glass.
The image formation is the same as it was in the previous experiment.
It debunks the idea that the image forms on the mirror's surface.
The image behind the mirror is the same distance as the object in front of it.
There is no real light behind the plane mirror that makes the image we see look real.
The reflected light appears to come from a point behind the mirror.
You place a lamp in front of a mirror and tilt it so that the can see from the diagram below, the image of each top and bottom of the lamp are at different distances behind the mirror.
Where do you see the image of the lamp?
The virtual image of the lamp is produced by the mirror.
The virtual image of each point of an object appears to come from a tilted image behind the mirror, which is reflected light from the top and bottom of the tilted lamp.
You can find the images of the end age quadruples by looking at the distance between the observer and the lamp.
When buying a mirror, you only need to see yourself from head to toe.
You can draw a sketch of the situation.
If the rays from the top of your eyes reach your eyes after reflection, you will see your toes.
If ray 2 from the top of your head reaches your eyes, you will see the top of your head.
You can still see your whole body by using the mirror.
If you mounted the mirror on a vertical wall, all parts of your body should be in the mirror.
The vertical distance is between the top of your head and your eyes.
A mirror is on a wall.
The curved mirrors are made from a piece of glass backed by a metal film.
The paper shows us the path of the laser light reflected from the mirror.
A diagram of the plane of the page.
The rays are reflected to the mirror's surface.
The law of reflection is used to analyze the principal axis.
The paths of the light are the same as reflected beams.
The lines here are the same as on the principal axis.
We use the law of reflection to draw a diagram but not parallel to the axis.
The path of the light is one ray.
The rays reflect back at the same point as they pass through the center of C.
The diagram matches the results of the experiment.
According to the law of reflection, the mirror reflects light.
The normal line goes through the center of the mirror.
The focal point of the mirror is where incident rays travel parallel to the principal axis.
All incident rays traveling parallel to each other but not parallel to the principal axis intersect at a common point after reflection.
The focal point is on the line that is parallel to the principal axis.
A curved mirror reflects light in agreement with the law of reflection.
The incident rays can be compared to the mirror on parallel rays.
They can pass through the focal point if they are not paral el to the principal axis.
The plane that passes through the focal point is called the focal plane.
The rays converge after reflection.
We can easily determine the location of the focal point and pass through a point in the focal plane if we consider the rays of the Sun to be paral el.
The method is used in Reasoning Skill 22.1.
The line goes through the mirror.
We will verify this assumption later.
The object can be represented by two or three rays.
Light is emitted in all directions.
The rays are directed after the tip of the object arrow.
The observer can see the image.
It travels parallel to the axis after reflection.
The mirror reflects back in the same direction after hitting it.
If you place the screen closer to the mirror than the image location, the reflected light will reach it but the image will be fuzzy because the rays pass through different points on the screen.
The law of reflection states that each point on an object emits an infinite number of rays that reflect off the mirror.
The Reasoning Skil was chosen because we know how they re flect without having to draw the normal line to the point of incidence and measure the angles of incidence and reflection.
We used three special rays to locate the image of an ob near the ject in In Reasoning Skil 22.1 Ray 1 traveled from the tip of the object to the mirror.
The mirror paral el is affected by the rays.
The virtual image appears to be the focal point after reflection.
Ray 2 went through the focal point.
The three rays are used in our study of mirrors.
Real images are not all reflected rays.
rays 1 and 2 will be used to build a ray dia gram.
The light seems to come from a point behind the mirror for a person standing behind it.
The image is larger than the object.
You can see that image when you use a mirror.
The image of the top of the object has been included in our ray diagrams so far.
Locating the base of the ray that we use to locate the image of the base will have to intersect the prin image produced by a curved mirror.
A line parallel to Ray 2 goes through the center of curvature.
If you draw a ray 2 from the base of the object and plane, it will go through the same point as the focal reflected ray goes through that plane.
The image of the base is located here.
The image of the top of the arrow is not shown in the figure.
Let's look at mirrors.
The reflected beam goes through the normal lines.
Light rays moving parallel to the main axis of a mirror reflect and move away from each other.
A mirror shows rays parallel to the axis.
The reflected rays go through the rays parallel to the principal axis.
They are different from the focal point behind the mirror.
The tip of the arrow is represented by two rays from the principal axis.
The observer sees something.
The rays should be behind the mirror.
The mirror reflects back in the opposite direction after hitting it.
A diagram of a ray.
Reasoning Skill 22.2 describes the types of rays 1 and 2 that can be used to locate images of an object.
The object should be placed behind the mirror to be smaller than the object.
The images are smaller than the object and the same size as mirrors are always upright.
You have found a mirror.
The distance from the mirror to the image depends on the mirror equation.
The points on the mirror are M and N. The rays are close to the principal axis and the mirror isn't curved very much.
We make the steps to complete the derivation using this method.
If the mirror is not very curved, you can use #AB and #ND A1B MD1.
The ABF and NDF are similar.
Both A1B1F and MDF are similar.
To get from the book.
The multiplication is done on the left side of the equation.
If the equation is consistent with an extreme case, we should test it.
We know that a mirror causes rays to go through the focal point after reflection.
We can assume that the rays from the object reaching the mirror are parallel to the principal axis if the object is far away.
The Eq is being used.
The image is at a distance from the number to the mirror.
The result is zero, it is consistent with the prediction.
The image of the tip of the mirror's curved surface should be found by using a ray diagram.
You can draw a sketch of the situ ation.
The distances can be easily thirds the object distance if you assemble all parts of the ex.
Place the screen here to see the image.
From the ray dia gram, we can see that the image is inverted and closer to the mirror than the candle.
The candle is 0.20 m from the mirror.
The negative sign means the image diagram.
The predictions are virtual.
The image was not real.
When using Eq, we need to agree on some sign conventions.
Let's see if the sign conventions work when we use the mirror equation for a mirror.
Below is a sketch of the situation.
The focal length is a negative number.
The number is (1>-5.67 m)-1
The ray diagram shows that the image distance is negative and its mag nitude is less than the object distance.
When the mirror is 1.0 m from her face, the image of your friend's face in a con vex mirror is upright, virtual, and 0.30 m behind the mirror.
The mirror was cut from a sphere.
The focal length is the object.
The prediction was minus 86 m.
If appropriate sign conventions are used, it appears that the mirror equation works the same for both types of mirrors.
The focal length is either positive or negative for the mirror.
There is a positive and a negative image distance.
Sometimes the size of an image produced by a curved mirror is larger than the size of an ob ject.
If the image is upright and inverted, the height is positive.
Since the object and image heights are not known, it is not possible to calculate magnification above.
The distance is determined by determining the linear image distance.
Use rays 1 and 2 in the ject.
The image will be inverted if it is real.
The inverted image's height is considered negative.
The ratios have different signs.
The neg ative sign is added to account for this.
The image is larger, upright, and virtual.
Reasoning Skil 22.1 states that you should use a mirror with a radius of 0.25 m for putting on makeup or shaving.
It doesn't hit the mirror.
The diagram shown below is a ray diagram.
There is a Rearranging Eq.
The birthmark image is larger than the object.
You hold the coin from the arrow.
The reflected mirror has a focal length of +0.60 m.
You place a mirror on a stand so that it is vertical and then slowly move a candle closer to the mirror from a large distance away.
The size of the mirror does not affect the magnification.
Its focal length does.
The size of the image will not change if you cut it in half.
The spherical surfaces are used to make the lens.
We can use a lens to create images.
A segment of sphere 2 images of objects.
The part pointed inward.
In this book, we only look at the two surfaces of spheres.
They are easier to analyze, so we will start with them.
The index of refraction of glass is greater than that of air.
When a ray of light moves from air to glass, it bends toward the normal line when it leaves the lens.
The ray bends toward the axis.
There are lasers passing through a lens.
After passing through the lens, the rays travel through the cipal axis of the glass lens.
The law of refraction is used to analyze the rays.
The boundary of the two media is not affected by Ray 2.
The rays travel through a point on the plane at an angle to the principal axis that passes through the focal point lens.
The law of refraction is used to analyze the rays.
The rays passing through the center of the lens do not bend.
The focal point of a curved mirror is similar to this point.
This plane is the focal plane.
A thick lens bends as it passes through a lens.
The focal point is closer to a lens light than a lens with less curvature.
There are two focal points on each side of the lens.
In our experiments, we shine light from the left side of the lens.
The rays will converge on the left side of the lens if we repeat the experiment.
We have used only glass.
A round bottle filled with water can act as a lens.
The beams converge at a point about 5 cm from the bottle after passing through it.
The bottle has a focal length of about 5 cm.
The focal len Normal gth of a lens depends on a number of factors, not the least of which is the material between the surfaces and the outside matter.
The longer the focal length of the lens, the less it bends light rays.
The focal point of the beams is different after passing through the lens.
The dashed lines show that the beams seem to go in different directions.
The lens is used in many devices.
The diagrams help us understand how these devices work.
Ray 3 moves parallel to the axis on the right by passing through the focal point on the left.
Both steps are the same.
Ray 1 is moving away from the principal axis.
The focal point on the left side of the lens is where Ray 1 appears to come from.
Ray 2 goes directly through the middle of the lens.
The focal point of Ray 3 is on the right.
The axis on the right is parallel to the ray.
The path of each ray can be explained by an image.
We need an image that is relative to the normal lines to the air-glass interface.
In a magnifying glass, we want the distance from the lens to the rays to be the same.
We will use the rays that were described in Experiment 2 in Table 22.5 to check the assumption.
Locating the image of the object's base.
The passes through the center of the image are parallel to rays 1 and 2.
An infinite number of rays leave the object.
Ray 1, paral el to the principal axis, does not bend as it passes through the center of the lens.
An imaginary ray 3 can be drawn parallel to ray 2 passing through the center of the lens.
There is an image of the object where bent rays 2 and 1 intersect on the principal axis.
The image of the arrow can be drawn using the top and bottom points.
The center of the lens does not bend as the image of a shining point-like object S is produced.
The rays also pass by a lens.
The axis through the image point is the line.
We draw a straight line.
Where is the lens, what kind of lens is it, and from the object to the image.
The image is on the opposite side of the lens from the object.
An unbent ray can be drawn from the object to the image to locate the lens.
The axis is at the position of the lens.
The shining point of the image S is located at the location of the lens that produced it.
The rays that reach the lens bend.
We need to find rays that are parallel to the principal axis.
The lens and its focal points will be affected by this ray.
The focal is shown to the left.
The other focal point of the lens is on the right.
A ray from the object that passes through the focal point on the left side of the lens should move parallel to the axis on the right side of the lens.
Imagine that you have a screen, a lens, and an object.
Half of the object is rays from a point on the object that is above the principal axis.
To answer the question, we need to draw a diagram and see what happens.
The top part of the lens is covered by an image brightness image.
The object is the lens.
The same technique was used for curved mirrors.
The results apply to con cave lens as well as the derivation.
We will assume that the radii of the spheres from which the sides of the lens were cut are larger than the size of the lens.
The lenses can be thin.
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We assume that the lens is very thin and that the rays are close to the axis.
Triangles B1A1O and BAO are similar.
The Triangles NOF and A1B1F are similar.
When we know the location of the object and the focal length of the lens, we can use the thin lens equation to predict the location of the image.
Let's see if this works in extreme cases.
An object is located at the focal point.
The rays from this object can be seen through the lens.
No image will be formed because they neither converge nor diverge.
It's equivalent to saying that no image forms.
The formation of an image by a lens is summarized.
The thin lens equation uses several sign conventions.
The image on the screen is blurry because the light from the tip of the arrow is spread out.
Being able to see where a sharp image will appear is important for all optical instruments.
Here we look at their optical components.
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