Reflection and Mirrors Legacy Problem #18 Guided Solution
Problem*
Kerry Uss is studying the convex side of her soup spoon. She notices that her 3.8-cm tall nose appears to be 1.2 cm tall when positioned a distance of 2.4 cm from the spoon.
- Determine the image distance for this particular object distance.
- Determine the focal length of the convex side of the spoon.
Audio Guided Solution
A good student of physics will blend effective problem-solving habits, some good thinking, and a solid understanding of physics concepts and equations in order to approach problems successfully. In this question, we have a student who places their nose in front of the convex side of a soup spoon. It acts as a reflector, creating an image of the nose. The nose is thus the object, and its height is 3.8 centimeters. I would call that HO, HO equals 3.8 centimeters. And its appearance of the nose, as viewed in the soup spoon, is 1.2 centimeters tall. That is the HI, the image height, HI equals 1.2 centimeters. It's a positive 1.2 centimeters, since it's an upright image, as formed by convex mirror. The distance of the nose to the spoon is 2.4 centimeters. That's DO, DO equals 2.4. I wish to calculate the image distance, DI, and the focal length, f. In order to calculate the image distance, I use the relationship that HI per hole ratio is equal to the negative of the DI per doe ratio. Rearranging that equation to solve for DI, I would have DI equal the negative of DO times HI divided by HO. That would be negative 2.4 times 1.2 divided by 3.8. And that comes out to be negative 0.75789, and I can round that to two significant digits, such that it's negative 0.76 centimeters. And that is the image distance. Now that I have image distance, along with the given object distance, I can find the focal length, which is the subject of Part B. I would use the mirror equation for this. The mirror equation states that 1 over f equals 1 over DO plus 1 over DI. The DO value is the given 2.4 centimeters, and the DI value is what was calculated in Part A, negative 0.75789. Substituting and evaluating the right side of the equation yields negative 0.90278 as being equal to 1 over f. Taking the reciprocal of each side would yield the value of f. It comes out to be negative 1.1077 centimeters. I can round that to two significant digits, such that it's negative 1.1 centimeters. The negative is consistent with the idea that it's a convex reflector.
Solution
- di = -0.76 cm
- f = -1.1 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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