A 1.80 - m - tall person stands 9.00 m in front of a large,...

Name: Problem 14, Chapter 23: College Physics
Uploaded: 2020-01-03T20:29:11-08:00
Duration: 3 min 18 s
Channel: Zachary Warner
Description: A 1.80 - m - tall person stands 9.00 m in front of a large, concave spherical mirror having a radius of curvature of 5.00 m. Determine (a) the mirror’s focal length, (b) the image distance, and (c) the magnification. (d) Is the image real or virtual? (e) Is the image upright or inverted?

A 1.80 - m - tall person stands 9.00 m in front of a large, concave spherical mirror having a radius of curvature of 5.00 m. Determine (a) the mirror’s focal length, (b) the image distance, and (c) the magnification. (d) Is the image real or virtual? (e) Is the image upright or inverted?

Key Concepts

Real vs Virtual Images

The real versus virtual image concept distinguishes images based on how light behaves. Real images are formed when light rays converge and can be projected on a screen; they are typically inverted. Virtual images, on the other hand, occur when light rays diverge, making the image appear to be located behind the mirror, and these images are usually upright.

Magnification

Magnification in mirror problems is defined as the ratio of the image height to the object height, and it can also be expressed as m = -di/do. This concept not only tells how the size of the image compares to the size of the object but also indicates the orientation (upright or inverted) of the image based on the sign of m.

Radius of Curvature

The radius of curvature is the distance from the mirror's surface to the center of the sphere from which the mirror segment is derived. In mirror problems, it directly relates to the focal length, as the focal length is typically half the radius of curvature (f = R/2).

Concave Spherical Mirrors

Concave spherical mirrors have a reflective surface that curves inward like the interior of a sphere. They have the property of converging light rays to a focal point and are commonly used in applications such as telescopes and headlights because of their image-forming abilities.

Mirror Equation

The mirror equation, given by 1/f = 1/do + 1/di, provides a relationship between the focal length (f), the object distance (do), and the image distance (di). It is fundamental for solving problems involving image formation in mirrors, enabling one to compute unknown quantities when two of the three are known.

Transcript

00:01 We are told that we have a mirror with a radius of curvature r of five meters and it's concave, okay, and that there is a person that is standing a distance p of nine meters from that mirror.

00:14 And we are asked to determine the focal length, the image distance, the magnification, and whether or not the image is real or virtual or upright or inverted.

00:24 So for part a, we can just use this information here to first find the, focal length, and all we need is the radius of curvature.

00:34 So that's because the focal length is equal to one -half the radius of curvature or r over two.

00:41 Therefore the focal length is simply 2 .5 meters and we can box that in as our solution for part a part b has asked us to determine the distance to the image so we can use the relationship between focal length image distance and object distance to find that.

01:01 So 1 over p plus 1 over q.

01:05 Q is what we're trying to find here.

01:06 That would be the image distance is equal to 1 over the focal length.

01:11 Or in other words, q is equal to negative the focal length times the period divided by the focal length minus the period.

01:26 So plugging those values into this expression, we find that this is equal to 3 .46 meters.

01:33 We can box that in as our solution for part b.

01:38 Now that we know the distance to the image as well as the distance to the object, we can go ahead and find the magnification for part c.

01:46 The magnification will call m is equal to negative q over p...