A dedicated sports car enthusiast polishes the inside and...

A dedicated sports car enthusiast polishes the inside and outside surfaces of a hubcap that is a section of a sphere. When he looks into one side of the hubcap, he sees an image of his face 30.0 cm in back of it. He then turns the hubcap over, keeping it the same distance from his face. He now sees an image of his face 10.0 cm in back of the hubcap. (a) How far is his face from the hubcap? (b) What is the magnitude of the radius of curvature of the hubcap?

Key Concepts

Mirror Equation

The mirror equation, 1/p + 1/q = 1/f, is central to problems involving image formation by spherical mirrors. It relates the object distance (p), the image distance (q), and the focal length (f) of the mirror. This equation is fundamental in optics and helps in predicting where an image will form based on the position of the object relative to the mirror's focal point.

Focal Length and Radius of Curvature Relationship

In spherical mirrors, the focal length is directly related to the radius of curvature (R) by the formula f = R/2. This relationship encapsulates the geometry of the mirror and is key to understanding how the curvature of a mirror impacts its focusing properties and image formation.

Sign Conventions for Spherical Mirrors

When working with spherical mirrors, it is crucial to adopt the appropriate sign conventions for distances and focal lengths. These conventions distinguish between real and virtual images and determine whether a mirror is concave or convex. Correctly assigning positive or negative signs to the object distance, image distance, and focal length guides the proper application of the mirror equation and ensures consistent results in calculations.

Image Formation by Spherical Mirrors

The production of images by spherical mirrors depends on the object's location relative to the mirror’s focal length. Depending on whether the mirror is concave or convex and whether the object is inside or outside the focal length, the mirror can produce real, inverted images or virtual, upright images. Understanding the conditions under which each type of image is formed is fundamental in analyzing and solving optical problems involving spherical mirrors.

Transcript

00:01 So we're given two situations when an individual is looking into a hubcap.

00:05 For one situation, he's looking at the hubcap and he sees his face 30 centimeters in back of it.

00:12 So q1 would be equal to 30 centimeters.

00:15 And then in the second one, he sees his image 10 centimeters in the back of the hubcap.

00:19 So q2 is equal to 10 centimeters.

00:22 Okay.

00:23 So now using that, it wants us to find for part a the distance the individual's faces away from the hubcap and for part b it wants us to find the magnitude of the radius of the hubcap.

00:36 Okay, so to do that, we're going to consider the fact that when the hubcap is convex in the first part, it'll have a specific relationship to the radius and when the con, when the hubcap is concave in the second part, it will have another relationship to the radius as well as the image distance.

00:59 So for the first part, for the concave part, one over negative q1, because he says the image is in the back of the hubcap, plus one over the distance to the object or the person's face is equal to two divided by the radius.

01:19 So that's for the first part or the convex part.

01:23 Now for the concave part, one over minus q2, again minus because he sees the image in the back of the hubcap, plus one over the distance to the individual's face, p, it's the same in both cases, is equal to 2 over r.

01:39 So now we have two equations and two unknowns.

01:44 Excuse me.

01:45 So for the first part, for the convex, since it's convex, the radius is negative here.

01:51 We can consider that to be a negative radius.

01:54 Okay, so we can add these two equations together.

01:59 So what we end up with is 1 over negative q1 plus 1 over p plus 1 over negative q2 plus 1 over p is equal to 0, right? because it's 2 over negative r plus 2 over r, so that's equal to 0.

02:26 Or in other words, trying to isolate p, 2 over p is equal to 1 over q1 plus 1 over q2.

02:46 Or in other words, solving for p, p is equal to 2 times q1 times q2 divided by q1 plus q2...

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