Reflection and Mirrors Legacy Problem #24 Guided Solution
Problem*
The real image produced by a concave mirror is observed to be six times larger than the object when the object is 34.2 cm in front of a mirror. Determine the radius of curvature of this mirror.
Audio Guided Solution
In this question we're told that an object is placed 34.2 centimeters in front of a mirror and we're told that an image is produced and it's a real image and it's six times larger than the object. What is important about this problem is that you'd be able to read that information and decipher what it's telling you. It's telling you that the ratio of high to whole is six but now the big question is, is it positive six or negative six? The difference between the plus and minus is the difference between wrong and right on this problem and the answer to is it positive or minus is that it's minus. Minus because the ratio of high to whole will be negative whenever you have an image which is inverted. Positive if you have an image which is upright. Real images are inverted and thus the high to whole ratio, the magnification, is negative six. Now that negative six, that high to whole ratio, is equal to the negative of the di to whole ratio and so I can say negative six equal negative di divided by the to whole value of thirty four point two centimeters. Doing so allows me to solve for the di value and when I do that I get 205.2 centimeters as my value of di. Now I'm asked to find the radius of curvature. In order to do that I need to first find the focal length and now having found di, I can find the focal length using the mirror equation. I just go one over F equal one over dou plus one over di and then I substitute thirty four point two in for dou and two hundred and five point two in for di. I evaluate the right side of that equation and I find that it's equal to 0.034113. That's equal to one over F. I wish to find the focal length and need to reciprocate that number. One divided by 0.034113 becomes 29.3143. That's the focal length and twice the focal length gives me the radius of curvature. That comes out to be fifty eight point six two eight six centimeters. I can round that to three significant digits such that the answer to this question is fifty eight point six centimeters.
Solution
58.6 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.