Refraction and Lenses Legacy Problem #14 Guided Solution
Problem*
Determine the critical angle of the following materials when surrounded by water (n = 1.33):
- Teflon (n = 1.38)
- Pyrex glass (n = 1.47)
- Polycarbonate glass (n = 1.59)
- Sapphire gemstone (n = 1.77)
- Diamond (n = 2.42)
Audio Guided Solution
The critical angle is the angle of incidence which causes the angle of refraction to be 90 degrees. When a ray of light approaches the boundary between two materials at the critical angle, the refracted ray would lie along the boundary 90 degrees from the normal to that boundary. Critical angles are only experienced when light is traveling through the more dense material and approaching the less dense material. So in this problem, the only instances in which you could ever have a critical angle would be instances in which the light is incident within a material which has an index of refraction greater than 1.33. To calculate the critical angle, we can use Snowe's Law. On one side of the equation, we're going to put the index of refraction of the more dense material and we're going to put the critical angle as the angle of incidence. That will be our unknown value. I usually call it theta critical or theta crit. On the right side of the equation, we're going to put the index of refraction of water and we're going to put the sine of 90 degrees. After all, if the critical angle is the angle of incidence which causes the refracted ray to be 90 degrees from the normal line, then we have to put the 90 degrees over on that side of the equation. Now in part A, we will go 1.38 times the sine of theta crit is equal to 1.33 times the sine of 90 degrees. The right side evaluates to 1.33. We can divide each side of the equation by 1.38. That will give us the sine of theta crit equal 1.33 divided by 1.38. We can then take the inverse sine or the second sine of both sides of the equation. That would mean second sine of 1.33 divided by 1.38. That's equal to the critical angle. When I do that on my calculator, I end up getting 74.5295 and I can round that to three significant digits, 74.5 degrees. On the other parts of this problem, we do much the same thing, only substituting 1.47 in the place of the angle of incidence or the index of refraction of the incident material. So in part B, it would be second sine or inverse sine of 1.33 divided by 1.47 and it repeats itself for the remaining parts of this question.
Solution
- 74.5°
- 64.8°
- 56.8°
- 48.7°
- 33.3°
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.
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