Reflection and Mirrors Legacy Problem #20 Guided Solution
Problem*
Baxter Nachure lives in the country along Sinewave road. It is difficult to pull out of the driveway onto the road since the road is curved and trees prevent him from seeing around the corner. He recently installed a large convex mirror at one of the curves to give him a wider angle of view. It has a focal length of -1.54 meters. Determine the magnification of an oncoming car located 35.8 m from the mirror.
Audio Guided Solution
A good problem solver not only has a solid understanding of physics concepts and mathematics, but is also able to employ effective problem solving habits. Those habits involve reading the problem carefully and identifying known quantities in terms of the variables within typical physics equations, and then plotting out a strategy as to how to get from the known quantities to the unknown quantities. Here in this question, we're told that a convex mirror is installed near a driveway. It has a focal length of negative 1.54 meters. It produces an image of an oncoming car that is located 35.8 meters away. We have an F equal to negative 1.54 meters, and we have a DO, or object distance, of 35.8 meters. What we wish to calculate is the magnification, the M value. Magnification can be found one of two ways, either as a high whole ratio, or as the negative didot ratio. Since there's no high whole information given in this problem, I'm going to have to proceed about finding the magnification by determining an image distance to go along with the object distance. Then I can apply the formula, magnification equal the negative di per do ratio. In order to get that image distance, I need to use the mirror equation along with the focal length, which is given as negative 1.54 meters, and the object distance of 35.8 meters. I rearrange that equation to the form of 1 over di, or 1 over F, minus 1 over object distance. Then I plug in negative 1.54 meters for F, and 35.8 meters for the object. When I do that, and I solve for the right side of the equation, I end up getting negative 0.67728. That's not the image distance, that's the reciprocal of the image distance. So if I take the reciprocal of this number, I get the image distance, and its value is negative 1.4765 meters. Now what I want to do is use that image distance in the magnification equation. I say M is equal to the negative di over do, or the negative of negative 1.4765, divided by 35.8. When I do that, I get a value of 0.041243 as the magnification, and I can round that to three significant digits.
Solution
0.0412
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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