Reflection and Mirrors Legacy Problem #19 Guided Solution
Problem*
A large spherical mirror sculpture is constructed in the town square at Physicston, Illinois. The sculpture consists of a large sphere with a diameter of 24 meters which is coated with a reflecting material. A 1.8-meter-tall photographer stands a distance of 38 m from the concave side of the sculpture and takes a picture. Determine the image distance and the magnification of the photographer.
Audio Guided Solution
In this problem, a photographer is standing 38 meters from a concave mirror. The mirror itself has a diameter of 24 meters. As such, its radius of curvature would be 12 meters, and its focal length would be one-half the radius of curvature, making the focal length 6 meters. We're told the photographer has a height of 1.8 meters. That would be HO equal 1.8 meters. What we wish to find is the image distance, di, in the magnification. So my strategy will be centered around using the mirror equation in order to determine the image distance. And then after I get the image distance, I'll use the equation magnification equal negative di over dou in order to calculate the magnification. So in the case of finding the image distance, I'm going to use the mirror equation and rearrange it to the form of 1 over di, the image distance, is equal to 1 over f, focal length, minus 1 over dou, the object distance. For dou, I'm going to put in 38 meters. And for f, I'm going to put in a positive 6 meters. So I go 1 divided by 6 minus 1 divided by 38, and I evaluate all that to be 0.14035. That's equal to 1 over di. So if I take the reciprocal of 1 over di, I get di. And if I take the reciprocal of 0.14035, I get 7.125 rounded to 7.1 meters. And that's the image distance. For the magnification, I'm going to take my image distance of 7.125, and I'm going to divide it by the 38 meters object distance, and then I'm going to insert a negative sign in front of it consistent with the equation. That comes out to be a magnification of negative 0.18750. In other words, the photographer's image is reduced in size compared to the size of the photographer himself or herself. And so I'm just going to round this to two significant digits, negative 0.19.
Solution
di = 7.1 m
Mag = -0.19
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!
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