Practice: A 4 cm tall object is placed in 15 cm front of a concave mirror with a focal length of 5 cm. Where is the image produced? Is this image real or virtual? Is it upright or inverted? What is the height of the image?

Subjects

Sections | |||
---|---|---|---|

Ray Nature Of Light | 11 mins | 0 completed | Learn |

Reflection Of Light | 12 mins | 0 completed | Learn Summary |

Refraction Of Light | 32 mins | 0 completed | Learn Summary |

Total Internal Reflection | 13 mins | 0 completed | Learn Summary |

Ray Diagrams For Mirrors | 36 mins | 0 completed | Learn |

Mirror Equation | 20 mins | 0 completed | Learn |

Refraction At Spherical Surfaces | 10 mins | 0 completed | Learn |

Ray Diagrams For Lenses | 23 mins | 0 completed | Learn |

Thin Lens And Lens Maker Equations | 25 mins | 0 completed | Learn |

Concept #1: Mirror Equation

**Transcript**

Hey guys, in this video we're going to talk about an equation for mirrors that will tell us pretty much everything that we need to know about the images produced by mirrors before we had to rely on ray diagrams which are the notoriously difficult to draw that are really only good for qualitative information with a mirror equation we'll be able to get definite quantitative information such as the image location, whether it's real or virtual, whether it's upright, inverted etc., OK let's get to it.What we're gonna focus on are spherical mirrors and that's basically all that you ever see. What a spherical mirror is is it's a mirror that's cut from a sphere. It's not really cut from a sphere these are shaped to exist as if they had been part of the sphere. So if you see the solid piece right here, the shaded-in piece, that's our actual mirror what I've gone ahead and done is drawn with dashed lines the sphere that this mirror appears to be part of and that sphere has a radius R. Now the mirror because it's a piece of this imaginary sphere that doesn't actually exist it's still defined by that radius and we call that for a curve we call it the radius of curvature that would be the radius of the sphere that that piece belonged to if it was part of the sphere. Now the focal length only depends upon this radius of curvature and it's just R divided by 2. Really easy to remember the focal length of a mirror, half the radius of its curvature once you know the focal length for any spherical mirror we have the mirror equation to determine where the image is located. We would say that 1 over the object distance how far the object is from the mirror plus 1 over the image distance how far the image is from the mirror equals 1 over the focal length of that mirror right which is just as we saw are over 2. In order to use the mirror equation properly there are some sign conventions that we need to memorize. A concave mirror which is a converging mirror right it's a mirror that when collimated light comes into it that collimated light eventually focuses on to a single point that converging mirror the concave mirror has a positive focal length. Always a positive focal length by convention and it can produce images that have a positive image distance or a negative image distance and we'll talk about what that means in a second. Convex mirrors on the other hand are diverging mirrors. When you have initially collimated light coming at it that light is so spread apart that light is never focused and by convention we assign a convex mirrors negative focal lengths and they can only produce negative image distances for plane mirrors which neither converge nor diverge light we say by convention that the focal length is infinity that's how we're going to use it in the mirror equation and it too can only produce negative image distances. So let's talk about these image distances and what that means if we have a positive image distance this image is real and inverted those two are always paired all real images are always inverted and they always have a positive image distance and if the image distance is negative this image is virtual and it's upright this always goes together all virtual images are always upright and they always have a negative image distance. So which of the three types of mirrors can produce a real image the only one that can produce a real image is a concave mirror and this should have seemed obvious a real image is only formed by a convergence of lights and the only mirror that converge light is the concave mirror. Any mirror that doesn't converge light cannot produce a real image so the other two types of mirrors are stuck producing virtual images. Lastly straight as a result of the mirror equation it's really easy to find the magnification of an image which is how tall it is relative to the object's heights and this is given by this equation right here in this orange box now the negative sign is a convention that may or may not be included in your textbooks and your lectures and any other resources you have what that negative sign is there for is to tell you whether or not the image is upright or inverted. If you have a positive magnification the images operate if you have a negative magnification the image is inverted. How do you get a positive magnification SI has to be negative. Which means the image has to be virtual. How do you get a negative magnification SI has to be positive which means the image has to be inverted OK sorry the image has to be real. Because that information is already given right here you don't actually need the negative sign of this equation it's just a convention but what the equation tells you is the size of the image relative to the object if the magnification is two the object is twice as tall the if the magnification is half the object is half as tall OK sorry the image if the magnification is two the image is twice as tall if the magnification is half the image is half as tall. Let's do a quick example. A 1.4 meter tall person stands 1 meter in front of a plane mirror. Where is the person's image located and how tall is it? So we have our miror equation which is always going to tell us where our image is located now the thing here is that the focal length is actually infinity and 1 divided by infinity is 0. What this means is if I moved the image distance to the other side we have that 1 over the image distance is the negative of 1 over the object distance or that the image distance is the negative of the object distance so it's -1 meter. A person stands 1 meter in front of a mirror so if the object distance is 1 meter the image distance is -1 meter. What does that mean about the image is it real or is it virtual? It's clearly virtual. And we knew this already that a plane mirror can only produce virtual images now how tall is it well to find the height we need to use the magnification equation. And I'll use the sign just to be proper about this. This is negative SI over SO which is negative 1 meter over 1 meter which is positive 1, the positive means that this image is upright. This is why the sign is pretty much relevant in the magnification equation we already knew that because this image distance was negative this image was virtual and therefore it had to be upright so we didn't gain any new information from the magnification or the sign of the magnification but a magnification of one means that the height of the image has to equal the height of the object which as we are told is 1.4 meters. This wraps up our discussion on the mirror equation. Alright guys thanks for watching.

Example #1: Object in Front of Convex Mirror

**Transcript**

Hey guys, let's do an example. A 5 centimeter object is placed 10 centimeters in front of a convex mirror if the radius of curvature of the mirror is 2 centimeters where is the image located? Is the image real or virtual? Is the image upright or inverted and what is the height of the image? Now the first three questions are all given to us by the image distance, the answers to those questions is given to us by the image distance right based on the image distance we know where the image is located obviously, we know whether it's real or virtual based on the sign and we know whether it's upright or inverted based on whether it's real or virtual. So really those first three questions are answered by the same piece of information alright. In order to find that image distance we need to use our mirror equation that 1 over the object distance plus 1 over the image distance is 1 over the focal length. Question is now what is the focal length? Well the focal length is going to be R over 2. In magnitude but since this is a convex near we know by convention the focal length has to be negative so I'm going to put a little negative sign in here so that I don't forget and I don't mess up the problem because I use the wrong sign. The radius of curvature is 2 centimeters so there's a -2 over 2 which is -1 centimeter now I can use the mirror equation so let me rewrite it to solve or sorry to isolate for the image distance this is one over S minus one over S not. This is going to be 1 over -1 minus 1 over 10 and if you want you can simplify this to use a least common denominator, this is -10 over 10 minus 1 over 10 which is -11 over 10 and then that makes finding the image distance as simple as just reciprocating the answer. This is where a lot of students make mistakes. You're finding the one over S I, you're not finding S I this is not the final answer. The reciprocal of it is the final answer and that is -0.91 centimeters. That's the image distance now is this image real or virtual well it's negative so it is virtual. Is it upright or inverted Well since it's virtual It has to be upright so this is virtual and upright those two have to go together and they always go with the negative image distance the only thing left is to find the height of the image which is given to us by the magnification so the magnification and I'm just going to drop the sign because the sign is frankly a waste of time we already know that it's upright so we don't care about the sign of the magnification the image distance is 0.91 the object distance is 10. So this is going to be 0.09 that's the magnification. That means that the image height is 0.09 times the object type. And the objects is 5 centimeters tall. So this is 0.45 centimeters. So if we want to sum up all the information about this image it is located 0.91 centimeters from the mirror technically behind the mirror it is a virtual image which is upright and has a height of 0.45 centimeters alright guys that wraps up this problem thanks for watching.

Practice: A 4 cm tall object is placed in 15 cm front of a concave mirror with a focal length of 5 cm. Where is the image produced? Is this image real or virtual? Is it upright or inverted? What is the height of the image?

Practice: You want to produce a mirror that can produce an upright image that would be twice as tall as the object when placed 5 cm in front of it. What shape should this mirror be? What radius of curvature should the mirror have?

0 of 4 completed

An upright object that is 2.0 mm tall is placed to the left of a concave spherical mirror that has focal length f = +20.0 cm. The image is 4.0 mm tall and is upright. Is the image real or virtual?
A. Real
B. Virtual

Which of the following best describes the image of a concave mirror when the object distance is equal to the radius of curvature?
a. virtual, upright and magnification one
b. real, inverted and magnification less than one
c. virtual, upright and magnification less than one
d. real, inverted and magnification one

An object is placed 6.0 cm to the left of a concave mirror. The image formed by the mirror is 8.0 cm to the right of the mirror. The image is
(a) real and inverted
(b) real and upright
(c) virtual and inverted
(d) virtual and upright

An object 6.0 mm tall is placed 0.45 m to the left of a spherical mirror. The image formed by the mirror is inverted and is 2.0 mm tall. What is the focal length of the mirror? Be sure to indicate whether the focal length is positive or negative.

You want to design a mirror for viewing erect images of objects. The image should have a magnification of 0.5 when the mirror is 2.0 m from an object. Find the shape and radius of curvature of the mirror.

Suppose you wanted to start a fire using sunlight and a mirror. Which of the following statements is most accurate?
a) It would be best to use a plane mirror.
b) It would be best to use a convex mirror.
c) It would be best to use a concave mirror, with the object to be ignited positioned halfway between the mirror and its center of curvature.
d) It would be best to use a concave mirror, with the object to be ignited positioned at the center of curvature of the mirror.
e) One cannot start a fire using a mirror, since mirrors form only virtual images.

A concave mirror is placed 0.800 m to the left of an object that is 2.00 mm tall. The mirror forms an image on a screen that is to the right of the object. The height of the image is 6.00 mm. What is the distance from the object to the screen?

An object located 125 cm from a concave mirror forms a real image 80 cm from the mirror. The mirror is then turned around so that its convex side faces the object, and is moved so that the image is now 40 cm behind the mirror. How far was the mirror moved?
1. 90.9091
2. 13.6957
3. 134.146
4. 97.2222
5. 70.0
6. 107.66
7. 164.434
8. 10.5263
9. 8.28829
10. 86.0377

An object with height y = 2 mm is placed a distance s = 24 cm in front of the vertex of a concave spherical mirror with radius of curvature R = 16 cm, as shown in the figure below. What are the image distance s' and the height y' of the resulting image?
s' = _______________
y' = _______________

As you walk away from a plane mirror on a wall, your image
A) is always a real image, no matter how far you are from the mirror.
B) may or may not get smaller, depending on where the observer is positioned.
C) changes from being a virtual image to a real image as you pass the focal point.
D) gets smaller.
E) is always the same size.

An erect object is 50 cm from a concave mirror of radius 60 cm. If the object is moved to a new position, such that the new lateral magnification is +2.5. The new object distance, in cm, is closest to:
A) 18
B) 24
C) 30
D) 36
E) 42

A production line inspector wants a mirror that produces an upright image with magnification of 6.6 when it is located 16.1 mm from a machine part. What kind of mirror would do this job?
1. convex mirror
2. Unable to determine.
3. concave mirror

An object is 17.9 cm from the surface of a reflective spherical Christmas-tree ornament 6.83 cm in radius. What is the apparent position of the image?

An object is 1.0 cm tall and its erect image is 4.0 cm tall. What is the exact magnification?

You have a concave spherical mirror with a 13.5 cm radius of curvature. You place an object on the mirror's axis, 16.3 cm in front of the mirror.How far is the object's image from the mirror?_________ cmIf it can be determined, is the image real or virtual?a) Realb) Cannot be determinedc) VirtualIf it can be determined, is the image upright or inverted with respect to the object? a) Uprightb) Cannot be determinedc) Inverted

Enter your friends' email addresses to invite them:

We invited your friends!

Join **thousands** of students and gain free access to **55 hours** of Physics videos that follow the topics **your textbook** covers.