Refraction and Lenses Legacy Problem #21 Guided Solution
Problem*
During an in-class exploration lab, students are looking through both converging and diverging lenses at various objects and observing the characteristics of the images. Moses becomes particularly intrigued by the diverging lens. He places it a distance of 4.8 cm from the lettering of his textbook and estimates that the image of the letters are one-fourth the size of the actual lettering.
- Calculate the image distance which results when Moses holds the lens this close to the textbook.
- Calculate the focal length of the lens.
Audio Guided Solution
A good problem solver always practices effective habits. Those habits involve reading the problem carefully and identifying the known and unknown quantities. In this question, the object happens to be the letters on a textbook, and they are viewed through a diverging lens. They are placed 4.8 cm from that textbook, and as such, the D object is 4.8 cm. The person looking at these letters estimates that the image of the letters are 1 fourth the size of the actual lettering. This gives us information about the relationship between image height and object height. It's like saying HI divided by HO equal 1 fourth. Now, of course, you always have to think, would it be positive 1 fourth or negative 1 fourth? And the answer to that question involves whether or not the image are upright or inverted. In this case, when you look at an object through a diverging lens, you get an upright image, and as such, the image height must be positive for a positive object height. So HI per HO, or image height divided by object height, equal positive 1 fourth. And the HI HO ratio is always equal to the negative of the DI DO ratio. So what we can say is that negative DI divided by DO is equal to positive 1 fourth. Since we know what DO is, the positive 1 fourth is equal to negative DI divided by 4.8 cm. We can solve for DI this way, and it comes out to be negative 1.2 cm. The negative on negative 1.2 cm simply indicates that our image is a virtual image, and located on the same side of the lens as the object itself. So now I know DI, the answer to part A, and I'm asked to calculate the focal length. The focal length of a lens is related to the object distance and the image distance by the equation 1 over f equal 1 over DO plus 1 over DI. I can substitute values of DO and DI into this equation, and it becomes 1 over focal length is equal to 1 over 4.8 plus 1 over negative 1.2. I can evaluate the right side of this equation using my calculator, or alternatively, I can find the lowest common denominator and evaluate the right side using fractions. When I evaluate this on my calculator, I find out that 1 divided by 4.8 plus 1 divided by negative 1.2 comes out to be negative 0.62500. If I take the reciprocal of this number, I get the focal length, and that comes out to be negative 1.6, and that's in units of centimeters.
Solution
- -1.2 cm (a - image distance indicates a virtual image)
- -1.6 cm (a - focal length indicates a diverging lens)
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 and record them in an organized manner. Equate given values to the symbols used to represent the corresponding quantity - e.g., \(\descriptive{d_o}{d_o,distance object} = 24.8\unit{cm}\); \(\descriptive{d_i}{d_i,distance image} = 16.7\unit{cm}\); \(\descriptive{f}{f,focal length} = \colorbox{gray}{Unknown}\).
- 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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