A concave spherical mirror has a radius of curvature of magnitude 24.0 cm . (a) Determine the ob

A concave spherical mirror has a radius of curvature of...

Key Concepts

Magnification

Magnification in mirror systems is defined as m = -di/do, where the magnitude of m indicates the size enlargement or reduction of the image relative to the object, and the sign indicates the orientation (negative for inverted and positive for upright images). This concept is critical when adjusting for desired image sizing effects.

Real vs. Virtual Images

Real images are formed when light rays actually converge, and they can be projected onto a screen, normally appearing inverted in concave mirrors. In contrast, virtual images are formed by the apparent divergence of rays and cannot be projected, usually appearing upright. Understanding the conditions under which each type arises is crucial for analyzing mirror problems.

Concave Spherical Mirrors

A concave spherical mirror is a curved reflective surface that bends incoming parallel light rays to a focal point. These mirrors can produce real or virtual images depending on the object's location relative to the focal point, and their properties are governed by the geometry of the sphere from which they are derived.

Mirror Equation

The mirror equation, expressed as 1/f = 1/do + 1/di, is fundamental in determining the relationship between the object distance (do), the image distance (di), and the focal length (f) of the mirror. This relationship allows one to calculate the position of the image given a particular object placement.

Ray Diagrams

Ray diagrams are schematic representations used to trace the paths of light rays as they reflect off mirrors. These diagrams help in visualizing the process of image formation, determining image location, size, orientation, and in distinguishing between real and virtual images.

Transcript

00:01 So we have a concave mirror and we're going to answer some questions about an image that is produced and where the position of the object would be in order to produce that image.

00:14 We're not given any information on the object placement or the image placement or we're given information on the radius of curvature of this concave mirror, which is 24 centimeters.

00:27 And we're also given that the magnification of the image is, i probably should put this somewhere else, it's three times as much as the object and it's upright.

00:51 So an upright image means that the magnification is positive, and three times that means it's three.

01:02 It says our magnification.

01:04 So usually in optics problems, we would start off by doing a ray diagram.

01:09 And trying to figure out where the image is.

01:13 But since we're given nothing on the object distance or the image distance, we're going to have to use some equations first.

01:24 And before we also use these equations, we need to determine our orientation for optics questions.

01:31 And we know that anything in front of the mirror is going to be positive.

01:39 So if the mirror is set at zero on a number line, anything to the left of that is going to be positive.

01:50 Anything to the right of this means going to be negative.

01:55 And that's the conventions for this book, this textbook.

02:00 Usually it might be in reverse.

02:02 This side might be positive, while anything to the left or in front of the mirror is negative.

02:08 But for this book, if we have an object that's in front of the mirror, that object distance, for example, the object distance is going to be positive.

02:22 And if the image is behind the mirror, for example, then the image distance is going to be negative.

02:30 So that's what we mean by these conventions here.

02:33 So i'll just leave the number line up for us to reference strength problem.

02:39 So the first thing we want to do is we want to use our magnification.

02:44 Well, actually, the first thing we want to do is we want to determine our focal length for this mirror.

02:48 The focal length for any curved mirror is f equals to r over 2.

02:56 R over 2 is 24 centimeters over 2, which equals to 12 centimeters.

03:11 So that is our focal length for this mirror.

03:15 And it's going to be positive because the radius of curvature, the radius of curvature is going to be set somewhere here.

03:23 And since it's set in front of the mirror, it's a positive length.

03:27 The positive length here.

03:31 And the radius of curvature is used to solve for the focal length.

03:39 And the focal length has the same sign as the radius of curvature.

03:43 So the radius of curvature might be here, for example, which is also positive.

03:59 So that's that.

04:02 So now we have our focal length.

04:04 We want to use the magnification equation to solve for a relationship between our object length, in our image length.

04:13 The magnification is given as minus q over p.

04:21 Q is the image length or the image distance and p is the object distance.

04:29 So if the magnification is plus 3, we can say that minus q over p is plus 3.

04:36 And we can also say that q is equal to minus 3p.

04:45 Okay, so that's another thing that we have to use in our question.

04:52 So we have that, we have our focal length, which is 12 centimeters, and we have our relationship between our object length, which is p, and our image distance, which is q.

05:03 Q equal to negative 3p.

05:10 So what we can do is we can use the mirror equation, and i'll use a blue for this, the, let's just say use mary equation here.

05:33 So what we're gonna do is we're gonna solve for either p or q.

05:44 And what that does is that since we know what the relationship between p and q is, we can substitute them, and then we can solve for that value.

05:54 We can solve for that variable, p or q.

05:56 So let's just write the mirror equation.

05:58 The mirror equation is 1 over p plus 1 over q equals to 1 over f and we already know what f is f is 12 centimeters a positive 12 centimeters so if we were to substitute our q for minus 3p we would get this we would get 1 over p minus 1 over 3p equals to 1 over 12 this would be centimeters so that's what we have so far.

06:49 So we could factor out a 1 over p we get 3 over 3 since 3 over 3 is 1 minus 1 over 3 that's the left hand side of the equation equals to 1 over 12 centimeters.

07:10 And the reason when we get 3 minus 1 over 3 is because you have when you factor out a 1 over p like this we get 1 minus 1 3rd, 1 minus 1 3rd.

07:26 And 1 is equal to 3 over 3.

07:30 That allows us to have the same common denominator here.

07:37 So what this ends up getting, or becoming, it ends up becoming 1 over p times 2 thirds equals to 1 over 12 centimeters.

08:01 So 1 over p is equal to 3 over 2 times 1 over 12.

08:14 And then you say well 3 over 12 is just 1 over 4 in the denominator so we can cross that out we can say this is 4 and 4 times 2 is 8 so we get 1 over 8 so p is equal to 8 centimeters that's our object distance p is our object distance and it's positive 8 centimeters so since our object is 8 centimeters positive.

09:04 Our object would be placed somewhere on this side of the mirror, since that's the positive side of the mirror.

09:13 And 8 centimeters is smaller than our focal length.

09:17 So if our focal length is this dot here, which we should label.

09:22 Let me see if we can label this here.

09:24 R is bigger than f.

09:29 It's two times more than f.

09:32 So r is out here and f is here.

09:35 Our image, which is eight centimeters or sorry our object which is positive eight centimeters is less than focal length which we found here to be 12 centimeters so it's just four centimeters less so it would be like right here i think that's a little weird so let me try that again i'll just draw a line or an arrow here so that's our object i'll put oh or i'll just write down object here so that's our object that's where a we would accurately place it since it's 8 centimeters, while our focal length is 12 centimeters.

10:46 So it would be closer to 0.

10:48 Our object length is closer to 0, which means it's closer to the mirror here.

10:54 If we are our mirror were to be placed at 0 on the number line.

10:59 So we have our object distance...

Please give Ace some feedback