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### Reflection & Mirrors

• When light hits a smooth reflecting surface, the angle of incidence equals the angle of reflection.
• Reflection can be used to form images.
• Reflected rays pass through the image for real images. If the reflected rays don't pass through the image, you have a virtual image.
• All virtual images are upright and have a negative image distance. Real images are inverted.
• Plane mirrors have no magnification, as the image distance and object distance have the same magnitude.
• The focal length of a spherical mirror is equal to half the radius of curvature (f=R/2).
• The mirror equation states that 1/f=1/do+1/di

### Reflection & Mirrors

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

• Intro 0:00
• Objectives 0:10
• Waves at Boundaries 0:37
• Reflected
• Transmitted
• Absorbed
• Law of Reflection 0:58
• The Angle of Incidence is Equal to the Angle of Reflection
• They Are Both Measured From a Line Perpendicular, or Normal, to the Reflecting Surface
• Types of Reflection 1:54
• Diffuse Reflection
• Specular Reflection
• Example 1: Specular Reflection 2:24
• Mirrors 3:20
• Light Rays From the Object Reach the Plane Mirror and Are Reflected to the Observer
• Virtual Image
• Magnitude of Image Distance
• Plane Mirror Ray Tracing 4:15
• Object Distance
• Image Distance
• Magnification of Image
• Example 2: Plane Mirror Images 7:28
• Example 3: Image in a Plane Mirror 7:51
• Spherical Mirrors 8:10
• Inner Surface of a Spherical Mirror
• Outer Surface of a Spherical Mirror
• Focal Point of a Spherical Mirror
• Converging
• Diverging
• Concave (Converging) Spherical Mirrors 9:09
• Light Rays Coming Into a Mirror Parallel to the Principal Axis
• Light Rays Passing Through the Center of Curvature
• Light Rays From the Object Passing Directly Through the Focal Point
• Mirror Equation (Lens Equation) 12:06
• Object and Image Distances Are Positive on the Reflecting Side of the Mirror
• Formula
• Concave Mirror with Object Inside f 12:39
• Example 4: Concave Spherical Mirror 14:21
• Example 5: Image From a Concave Mirror 14:51
• Convex (Diverging) Spherical Mirrors 16:29
• Light Rays Coming Into a Mirror Parallel to the Principal Axis
• Light Rays Striking the Center of the Mirror
• Light Rays Never Converge on the Reflective Side of a Convex Mirror
• Convex Mirror Ray Tracing 17:07
• Example 6: Diverging Rays 19:12
• Example 7: Focal Length 19:28
• Example 8: Reflected Sonar Wave 19:53
• Example 9: Plane Mirror Image Distance 20:20
• Example 10: Image From a Concave Mirror 21:23
• Example 11: Converging Mirror Image Distance 23:09

### Transcription: Reflection & Mirrors

Hi folks. I am Dan Fullerton and I would like to welcome you back to Educator.com as today we continue our study of optics as we focus on reflection and mirrors.0000

Our objectives are going to be to determine the angle of reflection of light from a smooth surface utilizing the Law of Reflection, utilize ray tracing to determine whether an image from a plane or spherical mirror is real or virtual, upright or inverted, enlarged or reduced in size...0009

...and utilize the mirror equation to relate object distance, image distance, image size, object size, and focal length.0027

When we talk about waves hitting a boundary, three different events can occur: the wave can be reflected, the wave bounces off the boundary, it can be transmitted where it continues into that new medium, or it can be absorbed, where it is transferred into the boundary.0036

Sometimes you will even hear people talk about it being scattered, which is a special type of reflection.0051

Focusing on reflection -- the Law of Reflection states that the angle at which a wave strikes a smooth, reflective medium, the angle of incidence (θI) is equal to the angle at which it reflects off the medium, the angle of reflection or θR.0058

The simple version of the Law of Reflection is θI = θR or the angle of incidence equals the angle of reflection.0073

Now when you measure the angle of incidence and angle of reflection, they are always measured from the normal, a line perpendicular to the reflecting surface.0082

If we have an incident ray of light here and a reflected ray, the normal is the line that is drawn 90 degrees to the reflecting surface.0090

The incident angle is always measured to the normal and the reflected angle is always measured to the normal and in that case θI = θR, the Law of Reflection, which is really that simple.0098

Now we talk about types of reflection and sometimes we have rough surfaces and the reflecting light is reflected in a variety of directions.0115

That is diffuse reflection. Think of reflections on walls or off a piece of paper.0122

But if you have smooth surfaces that reflect light waves in a more regular fashion so that the reflected waves maintain their parallelism, then this is called specular reflection.0127

You see this in things like mirrors, which is where we are really going to focus a lot of our time today.0137

Specular reflection -- The diagram below represents a light ray striking the boundary between air and glass.0144

What is the angle between this light ray and its reflected ray?0150

Well, first thing we will do is we will measure the angle of incidence, which is always measured to the normal.0154

We will take our ray, we will draw our angle (θI to the normal), and if that angle is 90 degrees, then this angle, θI, must be 60 degrees.0159

Our reflected angle must be the same thing, so as I draw this, I will do my best to gauge this correctly -- there is our reflected ray and we will call this θR reflected, which must also be 60 degrees.0169

So what is the angle between the light ray and its reflected ray?0186

Well, that is going to be this entire angle here, 60 + 60 or 120 degrees.0190

Now when we talk about mirrors, if you look at a flat or a plane mirror, you see reflections or an image of an object.0201

Light rays from the object reach the plane mirror and they are reflected to the observer and therefore you get an image.0207

The image is known as a virtual image though, because the reflected rays do not actually pass through the image.0213

The light rays hit the mirror, they come back to your eyes, but to you it seems like the image is behind the mirror, not where the light rays are actually going.0219

That makes it a virtual image; you cannot project a virtual image onto a screen, but a real image could be projected on a screen; a virtual image you cannot project.0227

All virtual images are upright and they have a negative image distance, which we are going to call (di).0238

Now the magnitude of the image distance is equal to the magnitude of the object distance in a plane mirror so the image appears the same size as the object.0245

As we analyze a lot of these mirror and lens situations, we are going to use what is known as ray tracing, drawing geometric representations of what is happening to the light rays.0255

The distance from the object to the mirror is the object distance or (do), so if this is our object right here, we will label (do) as the distance from our mirror to our object.0266

The [--] distance from the image to the mirror is the image distance (di) and it is pretty obvious to see that what we are going to have -- well let us see if we cannot draw this.0283

We are going to have the same object distance as image distance, so let us try and draw that here.0293

If there is our plane mirror and our eye is on this side, our image distance and our image is probably going to look something like that.0304

Now if we were to do the ray tracing here, in order to see the image it has to come back to the eyeball, so what we are going to do is we are going to draw a line from our eye down to the top of our image and I will extend it a little bit further...0313

...and because this is a plane mirror, if we draw the normal we have the same angle of incidence as the angle of reflection.0336

This is going to come back right here.0345

The light ray from here must be coming up on to the mirror and back to the [eyeball].0354

This is the virtual ray, it does not actually go that way; the light ray comes and hits the mirror into the eyeball and for the bottom of the mirror here...0361

...well the way we are going to draw that one is just going to be slightly different, but we know that that has to appear as if it is coming from the bottom of the image to the [eyeball].0369

It should look probably something kind of like this.0378

I will draw that and we will continue this down and back to the bottom of our image...0383

...so this light ray is coming from the bottom or our object, bounces off the mirror, again, following the Law of Reflection and back to the eye.0403

Because the image is behind the plane mirror, we know that the object distance and the image distance are going to be the same as this is a virtual image...0412

...and we can find the magnification of our image using the formula m (magnification) equals minus the image distance over the object distance or the height of the image divided by the height of the object.0421

In this case, the object and image distance are going to be equal.0433

The image distance will be negative, so the magnification is going to be just +1.0436

If we get a negative magnification, that just means the object is inverted.0442

Let us take a look at another example.0449

A student stands 2 m in front of a vertical plane mirror. As the student walks toward the mirror, what happens to the image?0451

Well, the magnification in a plane mirror is always going to be 1, so we have the same magnification and it remains virtual -- C -- always has to be that way for a plane mirror or an image in the plane mirror.0457

Which diagram best shows image (i) which is formed by placing and object (o) in front of a plane mirror?0473

Well you are going to get that reflection in a plane mirror.0479

The only one of these that shows the same thing reflected about the mirror is Number 2.0482

Well, let us talk now about something a little bit more interesting, like spherical mirrors.0491

Spherical mirrors are shaped like spheres or portions of spheres.0495

The inner surface of a spherical concave mirror is reflective, so if you had a mirror that looks like this and the inside here in the green area is where it is reflective, that is a concave spherical mirror.0498

If instead you had a spherical mirror -- say that portion of a sphere -- and it is the outside there that is reflective in green, that is a convex mirror.0510

Now the focal point of a spherical mirror is half its radius of curvature, so the focal point is the radius of curvature divided by 2.0520

As we start talking about these mirrors, here are a couple of points.0528

Concave mirrors are also going to be known as converging mirrors, because it will allow the light rays to converge.0531

Convex mirrors will be known as diverging mirrors because the light rays will not converge; they will diverge.0540

Let us take a look at our first example of converging or concave spherical mirror.0548

Light rays coming into a mirror, parallel to the principal axis are reflected from the plane of the mirror and converge through the mirror's focal point.0553

So here we have an object, here is the concave mirror; we have a principal axis and here is the focal point of the mirror, which is half the radius of curvature.0562

Our radius of curvature -- if that is our focal point right there, our radius of curvature is probably right about there (R).0571

Now light rays coming into a mirror parallel to the principal axis -- so that is going to be light rays coming from our object parallel -- we will draw that in blue all the way in to our mirror -- are reflected from the plane of the mirror and converged through the mirror's focal point.0584

That is going to come back through the focal point of our mirror.0605

We will label this and circle it in blue, so that you can see which ray it aligns with.0614

We also have light rays passing through the center of curvature are reflected back to the center of curvature, so if we were to draw a line from our object through the center of curvature (R), we will have to continue that...0618

...we will assume this mirror kept going and it would go in this way and come back right along the same path.0638

That we will label in purple and there is our second ray that we know for concave or converging mirrors.0645

The third ray we know -- light rays from the object passing directly through the focal point are reflected back parallel to the principal axis...0652

...so if we start from our object and we go through the focal point like this, then that is going to be reflected back parallel to our principal axis.0659

It comes in this way and goes back that way.0682

Now where all of these rays converge, that is going to be where we get our image, so if this is our object going from the principal axis to where the light rays converge here, we are going to have an image -- well it is going to be right here.0684

As you can see, it is going to be reduced in size and it is going to be inverted.0700

We have the light rays actually going through our image and that makes it real and it is upside down, so it is inverted.0706

All right. A basic spherical analysis and I should probably label this one in red, so you can see which rays go with which rules.0715

Now we can use the mirror equation, also known as the lens equation to relate the focal distance, the object distance, and the image distance.0725

The object and image distances are positive on the reflecting side of the mirror and negative on the non-reflecting side.0733

In this formula is 1/F = 1/do (object distance) + 1/di (image distance).0739

One of my favorite physics teachers remembers this by saying 'If I do, I die'.0747

It is kind of morbid, but if it helps you remember the formula, excellent.0754

Let us take a look at a concave mirror with an object that is inside the focal point.0759

Now I am going to start by drawing a ray that is parallel to the principal axis and that should get reflected through the focal point.0764

We will draw that from my object in parallel and that gets reflected through the focal point, which should bring it back that way.0772

I could also draw the line from the focal point through the object and that gets reflected back parallel to my principal axis.0782

But you will notice now that these two reflected rays are diverging; they are never going to meet and we are not going to get an image that way.0799

So what we have to do is we have to continue these back behind the plane of the mirror to find where our virtual image would be.0807

I will continue that back a ways and we will do the same thing for our first ray, our blue ray.0814

And it looks like they would meet right here, so that means if we were to draw our image, they are coming together again at the point of our image and I will draw that right in there and that must be its head.0824

Now we have an upright virtual image.0841

Again, using ray tracing, because these two diverged, we have had to continue these back on the opposite side of the mirror to get our virtual image.0848

Example 4 -- Concave spherical mirror. An incident light ray travels parallel to the principal axis of a concave spherical mirror.0863

After reflecting from the mirror, the light ray will travel...?0871

If it is coming in parallel to the principal axis, it is always going to be reflected through the mirrors focal point.0875

Light rays parallel to the principal axis are reflected through a mirror's focal point is one of our rules for ray tracing.0882

Let us take a look at the image from a concave mirror here in another example, where the diagram shows an object located at point (P) one-quarter of a meter from the concave's spherical mirror with focal point (F).0891

The mirror's focal length is .1 m, so we know that F = 0.1 m and our object distance (do) is 0.25 m.0904

How does the image change as the object is moved from point (P) toward point (F)?0917

As (P) is getting closer to (F), let us think about this -- we have our equation 1/F = 1/do + 1/di.0923

By moving the object toward point (F), the object distance (do), is going to decrease, so (do) is going to get smaller.0933

Since the focal point (F) is fixed, then 1/F must also be fixed, therefore, the image distance (di) must increase, so the image's distance from the mirror must be increasing.0942

Finally, if we use the magnification equation, m = -di/do -- we know that we have (di) increasing and (do) decreasing, so the numerator is getting larger and the denominator is getting smaller, so we must be seeing an increase in the magnification.0955

The image's distance from the mirror is increasing and its magnification is also increasing.0978

Now we talk about diverging mirrors or convex spherical mirrors, which has the same basic rules, but slightly adjusted.0990

Light rays coming into a mirror parallel to the principal axis are reflected away from the principal axis on a virtual line connecting the point of contact with the mirror plane and the focal point on the opposite side of the mirror.0997

Light rays striking the center of the mirror are reflected back at the same angle and the light rays never converge on the reflected side of a convex mirror, so convex mirrors only produce virtual images that are upright and reduced in size.1010

Let us take a look. Here we have some convex mirror ray tracings.1028

We have our object, we have a convex mirror, and now this side over here on the left is the reflecting mirror and the focal point we have drawn over here on the right.1031

By our first rule, if we are to draw our first ray that is parallel to the principal axis -- that is going to come in this way -- let us extend that a little bit further.1040

When it hits the mirror, it is going to be reflected, but the direction of reflection to figure that out, I have to draw a virtual line from my focal point through the point where it intersects the mirror, so there is our first ray.1055

Now our second ray -- the easiest way to do this one is the one that comes and hits the center of the mirror is just going to be reflected back the exact same direction.1077

Let us see if we cannot draw that pretty accurately.1085

We will be coming this way and I will extend that and then it is going to come back at the exact same angle.1089

Hopefully that looks relatively close.1108

So then, how do you find the image? Well, these are diverging again, so we have to extend this ray back as well on to the other side.1112

It looks like they cross right about here, so our image will be right there.1125

Notice that it is upright; it is reduced and it is virtual because the light rays are not actually converging where the image is, so an upright, reduced, virtual image from our convex mirror of ray tracing.1133

Example 6 -- Diverging rays -- Which optical device causes parallel light rays to diverge?1154

Well, that is very easy because that is a convex mirror, which is why it is called a diverging mirror.1160

Example 7 -- A radius of curvature of a spherical mirror is 20 cm, so R = 20 cm.1167

If an object is set 15 cm from the mirror, what is the focal length of the mirror?1175

Focal length is just radius of curvature divided by 2, so that is 20 cm/2 = 10 cm.1179

Another wave question. A sonar wave is reflected from the ocean floor.1194

For which angles of incidence do the wave's angle of reflection equal its angle of incidence?1198

It is called the Law of Reflection for a reason; it always holds, so that works for all angles of incidence.1203

I like this problem because it illustrates that what we are talking about is not always specific to just optics.1210

Law of Reflection holds for other waves as well.1215

A problem about plane mirror image distance -- In the diagram below, a person stands 5 m from a plane mirror -- there is our plane mirror -- and the chair in front of the person sits 2 m from the mirror.1221

What is the distance between the person and the image he observed through the mirror?1233

First thing is to figure out where the image he observed of the chair is.1238

If we have the object right here, then we are going to have the image of the chair, which should be about 2 m, exactly 2 m, on this side of the mirror, so we should have a chair that looks kind of like this and there is our image of the chair.1244

How far then is the person from the image he observed?1268

That means we are looking for that distance, so that is 5 m + 2 m or 7 m from the image of the chair.1272

Let us take a look at a problem with an image from a concave mirror.1284

An object arrow is placed in front of a concave mirror having center of curvature (C) and focal point (F).1288

Which diagram best shows the location of point (I), the image of the tip of the object arrow and is the image real or virtual, upright or inverted?1294

Let us take a look here. In this first one we have the ray parallel to the principal axis coming back through the focal point.1303

We have the line from the center of curvature coming back that way, but I do not think that is going to work.1312

Let us take a look. We have a line from the center of curvature going through here at parallel, but that is not how that works, so it cannot be 1.1321

Or Number 2, the line from the focal point through the object and then back off the mirror -- that is not right, that cannot work.1329

Number 3 -- We have the line parallel to the principal axis from the image that goes through the focal point and that works.1338

We also have the line through the center of curvature and the object going back through the center of curvature and that works and then they are extended back on the opposite side of the mirror to give you an image, so our image would look like that.1346

Now that is looking pretty promising, but notice in this case, we are going to have a virtual image that is upright, so yes, that has to be our answer because if we look over here at 4...1359

...well that is just silly. The line from the center of curvature through the top of the object is going back at a goofy angle, so that cannot be right either, so 3 must be our correct answer: the image is virtual and it is upright.1375

Let us try one more. A candle is placed .24 m in front of a converging mirror that has a focal length of .12 m.1389

How far from the mirror is the image of the candle located?1397

Well, let us use our mirror equation, 1/F = 1/do + 1/di or 'If I do, I die', and we are looking for the image of the candle, so I am going to rearrange this to say 1/di = 1/F - 1/do and then substitute in my values.1401

1/di is 1/.12 (focal length) and 1/do = .24, therefore 1/.12 - 1/.24 = 4.17 inverse meters or meters to the -1...1423

...therefore (di) will be 1/4.17 inverse meters, which is going to be 0.24 m.1446

Hopefully that gets you started on mirrors, reflection, and ray tracing.1460

Thanks so much for your time and we will talk a little bit more about ray tracing as we explore lenses in our next lesson.1465

Make it a great day everyone!1471