Reflection and Mirrors Legacy Problem #26 Guided Solution
Problem*
A child at an amusement park stands in front of a concave mirror with a focal length of 73.9 cm. With great amusement, the child holds her cotton candy close to the mirror and observes that its image is magnified by a factor of five. Determine the object distance which creates this magnification of five.
Audio Guided Solution
There is a concave mirror at the amusement park and it has a focal length of 73.9 cm. I write down f equals 73.9 cm. With great amusement, a child holds their cotton candy close to the mirror and observes that its image is magnified by a factor of 5. I'm going to write down this statement based on that sentence. I'm going to write the height of the image equal 5 times the height of the object. The image height equal 5 times the object height. And then I'm going to write a positive in front. The high value has got to be positive if the cotton candy is placed close to the mirror. It produces a virtual upright image. So height equal 5 times height can be rearranged to height divided by height equal positive 5. And the height per height ratio is equal to the negative height per height ratio. And so this positive 5 is also equal to negative height over height. What I wish to solve for is height. In order to solve for height, you either need to know height in order to use the mirror equation, or you either need to know di in order to use the equation that positive 5 equals negative di over dou. We don't have either. We don't have di, so we can't use either equation to find dou. But what we can do is use both equations together in order to solve for the two unknowns, di and dou, with dou being the one we're really wanting to find. So I'm going to begin by taking the statement that positive 5 equals negative di over dou and multiplying both sides of that equation by dou. I would get di equals negative 5 over dou. Now what I have is I have an expression for di written in terms of dou, and I can substitute that into the mirror equation. So I'm going to write the mirror equation as 1 over focal length, or 73.9, is equal to 1 over dou plus 1 over di. And instead of doing 1 over di, I'm going to do 1 over negative 5 dou. After all, that's what di is equal to. So my equation is, and you might want to write it down, 1 divided by 73.9 equal 1 over dou plus 1 over negative 5 dou. And on the right side of this equation, there's two fractions, which will have to be added together. And the only way to add them together is to get a common denominator. And the common denominator I'm going to pick is the denominator 5 dou. So the first term, 1 over dou, becomes 5 over 5 dou if I multiply the numerator and denominator by 5. And then I can go minus 1 over 5 dou. That's just the previous plus 1 over negative 5 dou rearrange. So what I have is 5 over 5 dou minus 1 over 5 dou is equal to 1 over 73.9. Now the side with the two fractions can be combined by adding the numerators and putting them over the same common denominator. And that side becomes 4 over 5 dou. So 1 over 73.9 is equal to 4 over 5 dou. We can cross multiply, and I get 5 dou is equal to 295.6. We can divide through by 5, and I get dou is equal to 59.120. Then round the three significant digits, dou equal 59.1 centimeters.
Solution
59.1 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.
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