Refraction and Lenses Legacy Problem #16 Guided Solution
Problem*
During a demonstration of fiber optics, Mr. H uses an acrylic light pipe to demonstrate total internal reflection for light as it navigates its way through a curved medium, entering at one end and exitting at the other with very little loss of light along the way.
- If the acrylic has an index of refraction of 1.48, then determine its critical angle when surrounded by air.
- Determine the critical angle of the acrylic light pipe when surrounded by water (n=1.33).
Audio Guided Solution
This question pertains to a demonstration of total internal reflection. Total internal reflection of light occurs when light approaches a boundary at an angle of incidence greater than the critical angle, and when it's in the more dense material, approaching the less dense material. We can calculate the critical angle by applying its definition to Snowe's Law. The definition of the critical angle is the angle of incidence which causes an angle of refraction of 90 degrees. Here in Part A of this problem, we have to insert the unknown critical angle value into Snowe's Law on the same side of the equation as the 1.48. And then we would say 1.48 times the sine of theta critical is equal to 1.00 for air times the sine of 90 degrees. We can divide both sides of the equation by 1.48, giving us sine of theta critical equal 1.0 divided by 1.48. Then taking the second sine or inverse sine of both sides of the equation gives us the critical angle. When I do that here, second sine of 1.00 divided by 1.48, I get 42.5066. I can round that to three significant digits. In Part B of this problem, we're doing much the same thing, only we're doing it for the acrylic water boundary instead of the acrylic air boundary. So what changes is, on the right side of the equation, we use 1.33 times the sine of 90 degrees. The left side remains 1.48 times the sine of theta critical. Again, dividing each side of the equation by 1.48 and then taking the second sine of each side of the equation gives us the value of the critical angle, 63.98100. We can round that to three significant digits and it becomes 64.0 degrees.
Solution
- 42.5°
- 64.0°
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!
Get more information on the topic of Refraction and Lenses at The Physics Classroom Tutorial.