Reflection and Mirrors Legacy Problem #25 Guided Solution
Problem*
A shiny bauble (ornament) hangs on Mr. H's Christmas tree. The bauble has a radius of 4.8 cm. Matthew looks into the bauble and observes an image of his face which is one-eighth the size of his face. How far from the bauble is Matthew's face?
Audio Guided Solution
There are a range of difficulty levels on physics problems. There are those which are conceptually easy and mathematically difficult, those which are mathematically easy and conceptually difficult, and then there's those like this one, which is both mathematically difficult and conceptually difficult. It's a problem about a Christmas tree ornament that serves as a convex reflector. We're given the radius of the ornament. It's 4.8 centimeters. And as a convex reflector, it'll have a negative radius of curvature and a negative focal length. So the radius of curvature of the convex side of this reflector is negative 4.8 centimeters, and the focal length is always one-half that value, and it's negative 2.4 centimeters. A child looks into the ornament and notices that the image of his face is one-eighth the size of his face. The question is, how far is this child from the bauble? What is the object distance? Now, I think that you might want to write a few things down here because it's going to get quickly mathematically complicated. I'm going to begin with the statement that his image of his face is one-eighth the size of his face. Another way to write that mathematically is to say that the height, the h-i, the image height, is, or equal to, one-eighth times h-o, the size of his face. That is high equal one-eighth whole. It could be rearranged to say high per whole is equal to one-eighth. And that's the magnification. It's a positive magnification, as is the case with all convex mirrors. Now, if that's the high whole ratio, that's also the negative Didot ratio. So I'm going to rewrite this as negative Didot equal positive one-eighth. Now, I cannot solve for the Dot unless I have the Dot, or at least have an expression for Dot expressed in terms of Dot, or vice versa. So I'm going to take this relationship, negative Dot over Dot is equal to positive one-eighth. I'm going to cross-multiply. It comes out to be eight Dot equal negative Dot, or worded differently, Do equal negative eight Dot. Now, that's an expression for Do written in terms of Dot. And I can substitute that into the mirror equation for Do in order to solve for Dot. So I'm going to write the mirror equation as negative, or one divided by negative 2.4, is equal to one over Do plus one over Dot. I'm going to rewrite it again as one divided by negative 2.4 equal one over negative eight Di plus one over Dot. I've just substituted negative eight Di in for Do in this equation. Now, I have two fractions on the right side of this equation. In order to add fractions, you have to have a common denominator. So I'm going to rewrite this as negative one over eight Di plus eight over eight Di. Now I have a common denominator. I'm going to reorganize the right side of the equation by adding the two numerators and putting them over the same common denominator. So negative one plus eight comes to seven, and that's all over eight Di. So I have one over negative 2.4 is equal to seven over eight Di. I can cross multiply now and get eight Di is equal to 16.8 centimeters. I can divide through by eight, and I get Di equal negative 2.1 centimeters. Only problem is, I don't want to solve for Di. I want to solve for Do. But that's no big deal, because previously I've written Do equal negative eight Di. Substituting in negative 2.10 into that equation, I'm going to get Do equal 16.8 centimeters. And that's the answer to this question.
Solution
16.8 cm
Habbits of an Effective Problem Solver
- Read the problem carefully and develop a mental picture of the physical situation. If necessary, sketch a simple diagram of the physical situation to help you visualize it.
- Identify the known and unknown quantities in an organized manner. Equate given values to the symbols used to represent the corresponding quantity - e.g., \(\descriptive{d_o}{d_o,distance object} = 24.2\unit{cm}\); \(\descriptive{d_i}{d_i,distance image} = 16.8\unit{cm}\); \(\descriptive{f}{f,focal length} = \colorbox{gray}{Unknown}\).
- Use physics formulas and conceptual reasoning to plot a strategy for solving for the unknown quantity.
- Identify the appropriate formula(s) to use. Perform substitutions and algebraic manipulations in order to solve for the unknown quantity.
Read About It!
Get more information on the topic of Reflection and Mirrors at The Physics Classroom Tutorial.