Printable Resources
Bending Light
Appendix A: Day 1 – Pre/Post Test
Appendix B: Day 1 – Pre/Post Test Answer Key
Appendix C: Day 1 – Guided Research
Appendix D: Day 1 – Guided Research Answer Key
Appendix E: Day 3 – Pythagorean Theorem Student Handout
Appendix F: Day 3 – Trigonometry Student Handout
Appendix G: Day 3 – Pythagorean Theorem and Trig Worksheet
Appendix H: Day 3 – Pythagorean Theorem and Trig Worksheet Answer Key
Appendix I: Day 4 – Inquiry into Refraction
Appendix J: Day 4 – Inquiry into Refraction Teacher Notes
Appendix K: Day 4 – Inquiry into Refraction Rubric
Appendix L: Day 6 – The Engineering Design Challenge
Appendix M: Day 6 – The Engineering Design Challenge Rubric
Appendix N: Day 6 – The Daily Performance Rubric
Appendix O: Day 6 – The Engineering Design Process
Appendix P: Additional Teacher Resources
Appendix Q: Technical Brief
www.daytonregionalstemcenter.org
Appendix A: Day 1 – Pre/Post Test
Name _____________________________
Problems
Show all of work.
1. Solve the following equation for the variable b.
𝒂 ∗ 𝐭𝐚𝐧 𝜽 = 𝒃 ∗ 𝐭𝐚𝐧 𝜽
2. Solve for the variable a in the following equation when c = 14 and b = 10
𝒂𝟐 + 𝒃𝟐 = 𝒄𝟐
3. Using the right triangle below, find the value of x to the nearest tenth.
x
28 cm
45 cm
(Diagram not to scale)
4. Using the right triangle below, give the correct ratios for the sine, cosine and tangent of angle
C.
13
12
C
(Diagram not to scale)
Draft: 3/22/2016
Page 2
5. Using the right triangle below, calculate angle c to the nearest whole degree.
8
6
C
(Diagram not to scale)
6. Calculate the angle of refraction of light that is traveling through air and passes into water at
an angle of 50° from the normal.
7. You are a CSI (criminal science investigator) for the Miami Dade Police Department. A
suspect has been held for questioning for a recent home invasion. The perpetrator broke a
window to gain entry into the house. You have been given the evidence collected from the
suspect, which includes glass fragments. You also have been given a sample of the broken
glass from the crime scene and have determined it’s index of refraction to be 1.66. You
analyze the glass evidence from the suspect and find that an incident angle of 40° has an
angle of refraction of 22.78°. Was this suspect at the scene of the crime? Support your
answer!!
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Short Answer
8. List four applications for lasers
9. List the steps to the engineering design process.
Essay
10. When a straw is placed in a clear glass of water it appears to be “bent”. We know the water
is not physically bending the straw, so explain why it appears to be bent.
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Appendix B: Day 1 – Pre/Post Test Answer Key
1. 𝒂 ∗ 𝐭𝐚𝐧 𝜽 = 𝒃 ∗ 𝐭𝐚𝐧 𝜽
𝑏=
𝑎 ∗ tan 𝜃
tan 𝜃
Points
4
3
2
1
Rationale
The student correctly calculates the answer.
The student makes one error in calculating the answer.
The student makes more than one error in calculating the answer.
The student attempts to solve the problem but is unsuccessful.
2. 𝒂𝟐 + 𝒃𝟐 = 𝒄𝟐 ∴ 𝒂 = √𝒄𝟐 − 𝒃𝟐
𝑎 = √142 − 102 = 9.798
Points
4
3
2
1
Rationale
The student correctly calculates the answer.
The student makes one error in calculating the answer.
The student makes more than one error in calculating the answer.
The student attempts to solve the problem but is unsuccessful.
3. Answer:
𝑥 = 53.0𝑐𝑚
Solution:
𝑎2 + 𝑏 2 = 𝑐 2 Where 𝑎 = 45𝑐𝑚 𝑎𝑛𝑑 𝑏 = 28𝑐𝑚
(45𝑐𝑚)2 + (28𝑐𝑚)2 = 𝑥 2
2025𝑐𝑚2 + 784𝑐𝑚2 = 𝑥 2
So 𝑥 2 = 2809𝑐𝑚2 ⟹ 𝑥 = √2809𝑐𝑚2
𝑥 = 53.0𝑐𝑚
Points
4
3
2
1
Rationale
The student correctly calculates the answer.
The student makes one error in calculating the answer.
The student makes more than one error in calculating the answer.
The student attempts to solve the problem but is unsuccessful.
4. Answer:
12
𝑠𝑖𝑛 𝐶 =
𝑐𝑜𝑠 𝐶 =
13
Points
4
3
2
1
5
13
𝑡𝑎𝑛 𝐶 =
12
5
Rationale
The student correctly states all three ratios
The student makes one error in stating the ratios
The student makes more than one error in stating the ratios
The student attempts to solve the problem but is unsuccessful.
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5. Answer:
6
𝐶 = 𝑠𝑖𝑛−1 ( ) = 49°
8
Points
4
3
2
1
Rationale
The student correctly calculates the angle.
The student makes one error in calculating the angle.
The student makes more than one error in calculating the angle.
The student attempts to solve the problem but is unsuccessful.
6. Answer:
The angle of refraction is calculated using Snell’s Law:
𝑠𝑖𝑛 𝜃𝐼 ∗ 𝑛𝐼 = 𝑠𝑖𝑛 𝜃𝑅 ∗ 𝑛𝑅
𝑠𝑖𝑛(50°)
𝜃𝑅 = 𝑠𝑖𝑛−1 [
] = 35°
1.33
Points
4
3
2
1
Rationale
The student correctly calculates the angle of refraction by using Snell’s Law
The student makes one error in calculating the angle of refraction by using Snell’s
Law
The student makes more than one error in calculating the angle of refraction by
using Snell’s Law
The student attempts to solve the problem BUT with out using Snell’s Law
7. Answer
Using Snell’s Law you can determine the index of refraction of the glass from the suspect:
(Note the index of refraction for air is 1)
𝑛𝐼 𝑠𝑖𝑛 𝜃𝐼 = 𝑛𝑅 𝑠𝑖𝑛 𝜃𝑅
𝑛𝑅 =
𝑛𝐼 𝑠𝑖𝑛 𝜃𝐼
𝑠𝑖𝑛 𝜃𝑅
1∗𝑠𝑖𝑛(40°)
= 𝑠𝑖𝑛(22.78°) = 1.66
Although the glass from the suspect has the same index of refraction as the glass from the
crime scene, this alone is NOT enough to conclude that the suspect was at the crime scene.
It DOES indicate that more tests should be done on the glass. The results from this analysis
only indicate that the glass samples are the same type of glass.
Points
4
3
2
1
Rationale
The student correctly calculates the index of refraction AND states that it is NOT
enough to conclude the suspect was at the scene.
The student correctly calculates the index of refraction BUT states that it IS enough
to conclude the suspect was at the scene.
The student incorrectly calculates the index of refraction AND incorrectly supports
their rationale as to if the suspect was at the scene of the crime.
The student made an attempt but is incorrect.
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SHORT ANSWER
8. Answer
Answers may vary. Applications within the following groups could be used: industrial,
environmental, communications, research, medical.
Points
4
3
2
1
0
9. Answer:
Points
4
3
2
1
0
Rationale
Student listed four applications.
Student listed three applications.
Student listed two applications.
Student listed one application.
Student did not list any applications
Rationale
Student correctly identifies all the steps of the engineering design process. (Identify
problem, formulate question to be answered, think about possible solutions, design
prototype, test the prototype, redesign prototype)
Student omits 1 step from the process
Student omits 2-3 steps from the process
Student omits 4-5 steps from the process
Student cannot name any steps of the process
ESSAY
10. Answer
Answers will vary. An exemplar response is:
Light travels at different speeds in different mediums. In a vacuum, light travels at 3x10 8 m/s
compared to 2.25x108 m/s in water. For us to see the straw, light has to travel through two
different mediums (water and air). It is because the speed of light is different in these two
mediums the light is refracted. We can see this refraction because the straw appears to be
bent.
Points
4
3
2
1
Rationale
The student describes how light travels at different speeds in different mediums
AND uses the term “refracted” or “refraction” correctly.
The students describes how light travels at different speeds in different mediums
BUT uses the term “bend” or “bent” instead of “refracted” or “refraction”.
The student uses terms such as “medium” and “refracted” but incorrectly.
The student attempted an answer but it is incorrect.
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Appendix C: Day 1 – Guided Research
Name _____________________________
Before you begin your study of light, lasers and the bending of light, you need to get a
little background information and knowledge. Find the definitions of the following terms
and answers to the following questions using the Internet and/or a textbook. Write all
your answers on a separate sheet of paper.
Define the following terms:
1.
2.
3.
4.
5.
6.
7.
8.
9.
Reflection
Refraction
Incidence
Wavefront
Laser
Index of refraction
Speed
Velocity
Pythagorean Theorem
Answer the following questions:
1. Lasers have many applications. Find and describe five (5) applications for the use of
lasers.
2. Briefly describe why light will “bend” when passing from one medium to another.
3. Who was the early 17th century Dutch mathematician who is best known for writing
equation for the refraction of light? Briefly write about his discoveries and give the
equation that he is known for.
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Appendix D: Day 1 – Guided Research Answer Key
NOTE: The Guided Research is used to get the students accustomed to the vocabulary and
information they will be learning about in this STEM lesson. Therefore the grading of this
assignment is very subjective. As a suggestion, a student could receive a score of 20 points for
completing the entire assignment with most answers correct.
Define the Following Terms:
1. Reflection- The throwing back by a body or surface of light, heat, or sound without absorbing
it: "the reflection of light"
2. Refraction- Change in direction of propagation of any wave as a result of its traveling at
different speeds at different points along the wave front.
3. Incidence - The intersection of a line, or something moving in a straight line, such as a beam
of light, with a surface.
4. Wavefront - an imaginary surface joining all points in space that are reached at the same
instant by a wave propagating through a medium
5. Laser - an acronym for light amplification by stimulated emission of radiation; an optical
device that produces an intense monochromatic beam of coherent light.
6. Index of refraction - The index of refraction is the ratio of the speed an electromagnetic wave
in vacuum to the speed of the electromagnetic wave in a particular medium. Typically
denoted by the lowercase n
7. Speed - The rate at which someone or something is able to move or operate: "the car has a
top speed of 147 mph". Also distance travelled per unit time.
8. Velocity - The rate of change of displacement with respect to time.
9. Pythagorean Theorem A theorem attributed to Pythagoras that the square of the hypotenuse
of a right triangle is equal to the sum of the squares of the other two sides.
Answer the Following Questions:
10. Lasers are used for numerous applications. Research to locate and describe five (5) different
applications for the use of lasers.
Answers will vary. An example: Lasers are used in drilling and cutting, alignment and
guidance, and in surgery; the optical properties are exploited in holography, reading bar
codes, and in recording and playing compact discs.
11. Briefly describe why light “bends” when passing from one medium to another.
Refraction is the change in direction of propagation of a wave when the wave passes from
one medium into another, and changes its speed. Light waves are refracted when crossing
the boundary from one transparent medium into another because the speed of light is
different in different media. he bending occurs because the wave fronts do not travel as far in
one cycle in the glass as they do in air.
(http://electron6.phys.utk.edu/optics421/modules/m1/reflection_and_refraction.htm)
12. Write the equation for the refraction light and identify the early 17th century Dutch
mathematician best known for discovering this equation? Write a brief summary of his
discoveries.
Willebrord van Royen Snell l is known for “Snell’s Law” or the “Law of Refraction”.
𝒏𝒊 ∗ 𝒔𝒊𝒏(𝜽𝒊 ) = 𝒏𝒓 ∗ 𝒔𝒊𝒏(𝜽𝒓 )
Where 𝜽𝒊 ("theta i") = angle of incidence 𝜽𝒓 ("theta r") = angle of refraction
ni = index of refraction of the incident medium
nr = index of refraction of the refractive medium
(http://www.physicsclassroom.com/class/refrn/u14l2b.cfm)
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Appendix E: Day 3 – Pythagorean Theorem: Student Handout
Name _____________________________
If we take the length of the hypotenuse to be c and the length of the
legs to be a and b, then the Pythagorean Theorem tells us:
c
b
a2 + b2= c2
By definition, the Pythagorean Theorem is:
a
A theorem stating that the square of the length of the hypotenuse of a right triangle is equal to the
sum of the squares of the lengths of the other sides.
Note that this is only true for “right triangles” (triangles that have a 90° angle).
Let’s look at an example to see how we can use the Pythagorean Theorem to find the
hypotenuse:
Find the value of x to the nearest
tenth:
x
73 cm
27 cm
Solution:
𝑎2 + 𝑏 2 = 𝑐 2 Where 𝑎 = 27𝑐𝑚 𝒂𝒏𝒅 𝑏 = 73𝑐𝑚
(27𝑐𝑚)2 + (73𝑐𝑚)2 = 𝑥 2
729𝑐𝑚2 + 5329𝑐𝑚2 = 𝑥 2
So 𝑥 2 = 6058𝑐𝑚2 ⟹ 𝑥 = √6058𝑐𝑚2
𝑥 = 77.8𝑐𝑚
(Diagram not to scale)
We can also use the Pythagorean Theorem to find a missing side, if we are given the hypotenuse
and the other side. Here is an example:
Find the value of x to the nearest
tenth:
74 cm
x
21 cm
Solution:
𝑎2 + 𝑏 2 = 𝑐 2 Where 𝑎 = 21𝑐𝑚 𝒂𝒏𝒅 𝑐 = 74𝑐𝑚
(21𝑐𝑚)2 + 𝑥 2 = (74𝑐𝑚)2
𝑥 2 = 5476𝑐𝑚2 − 441𝑐𝑚2
So 𝑥 2 = 5035𝑐𝑚2 ⟹ 𝑥 = √5035𝑐𝑚2
𝑥 = 71.0𝑐𝑚
(Diagram not to scale)
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Appendix F: Day 3 – Trigonometry: Student Handout
Name _____________________________
Right triangles are special. So special that mathematicians studied them intently. They
discovered some fascinating properties that relate the sides (including the hypotenuse) and the
angles of right triangles. These properties are the foundations of what is called “Trigonometry”
(or “Trig” for short). We are going to focus on just three trigonometric functions: sine, cosine and
tangent.
We will begin by defining each function then we will show some examples of using them. We will
finish by showing how to calculate angles using inverse trig functions.
The Definitions: Given a right triangle with sides a, b and c and angles A and B we have:
sin 𝐵 =
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒 𝑏
=
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝑐
sin 𝐴 =
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒 𝑎
=
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝑐
cos 𝐵 =
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒 𝑎
=
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝑐
cos 𝐴 =
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒 𝑏
=
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝑐
tan 𝐵 =
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒 𝑏
=
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒 𝑎
tan 𝐴 =
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒 𝑎
=
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒 𝑏
A
c
b
B
a
Keep in mind that the trigonometric functions merely tell us the ratio of 2 sides of a right triangle.
Here are two examples of how to determine all 3 trigonometric relations:
Using the given triangle,
determine the following:
a. sin 𝜃
b. cos 𝜃
c. tan 𝜃
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
25𝑚
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
30𝑚
a. sin 𝜃 = ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 = 39.1𝑚 ≈ 0.640
39.1m
25m
θ
b. cos 𝜃 = ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 = 39.1𝑚 ≈ 0.768
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
25𝑚
c. tan 𝜃 = 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 = 30𝑚 ≈ 0.833
30m
(Diagram not to scale)
Using the triangle given,
determine the following:
a. sin 𝜃
b. cos 𝜃
c. tan 𝜃
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
10𝑚
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
98𝑚
a. sin 𝜃 = ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 = 98.5𝑚 ≈ 0.102
98.5m
10m
θ
b. cos 𝜃 = ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 = 98.5𝑚 ≈ 0.995
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
10𝑚
c. tan 𝜃 = 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 = 98𝑚 ≈ 0.102
98m
(Diagram not to scale)
Draft: 3/22/2016
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It might be necessary to employ the Pythagorean Theorem before you can find all the
trigonometric relations. Here is an example:
First you must find the Hypotenuse using
the Pythagorean Theorem:
Using the triangle given,
determine the following:
𝑎2 + 𝑏 2 = 𝑐 2 ∴ 𝑐 = √(90𝑚)2 + (26𝑚)2 ≈ 93.7𝑚
a. sin 𝜃
b. cos 𝜃
c. tan 𝜃
26m
Now continue using the Trig definitions:
θ
90m
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
26𝑚
a. sin 𝜃 = ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 = 93.7𝑚 ≈ 0.278
(Diagram not to scale)
b. cos 𝜃 =
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
=
90𝑚
93.7𝑚
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
26𝑚
≈ 0.961
c. tan 𝜃 = 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 = 90𝑚 ≈ 0.289
Recall that the trigonometric functions tell us what the ratio is between 2 sides of a right triangle.
This is nice, but what if we want to know what angle gives us this ratio? If we know sinθ = 0.5
what does θ equal? Enter the inverse trigonometric functions! The inverse trig functions (also
known as the “arc” trig functions) will give us an angle for a known ratio.
For example, in the last exercise we found that sin 𝜃 = 0.278, if we employ the inverse sine
function we will get the angle: 𝜃 = sin−1 (0.278) = 16.1°
What if we had used the inverse cosine or inverse tangent? We would have gotten the same
angle!! Go ahead; try it!
Notice the -1 exponent? That is used to show we are using the “inverse sine” function. So in a
nutshell, sin 𝜃 gives a ratio and sin−1 𝑥 gives an angle.
A Few Calculator Tips Before You Go:
The trig functions are usually listed on a “button” as “sin”, “cos” and “tan”. To get to the inverse
functions, you will generally have to press another button then press a trig button. On a TI83/84+
that button is the “2nd” button. Also, angles can be measured in degrees and radians. Your
calculator should have a “mode” button where you can see (and change) what angle mode the
calculator is in.
Draft: 3/22/2016
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Appendix G: Day 3 – Pythagorean Theorem and Trig: Student Handout
Name _____________________________
Pythagorean Theorem and Trig Functions
Show all your work!! If you don’t have enough room, use another piece of paper.
PROBLEMS
1. Find the value of x to the nearest tenth:
x
28 cm
45 cm
(Diagram not to scale)
2. Find the value of x to the nearest tenth:
x
67km
30km
(Diagram not to scale)
3. Find the value of x to the nearest tenth:
x
85mm
87mm
(Diagram not to scale)
Draft: 3/22/2016
Page 13
4. Find the value of x to the nearest tenth:
95cm
x
74cm
(Diagram not to scale)
5. Find the value of x to the nearest tenth:
53km
x
20km
(Diagram not to scale)
6. Given 𝜃 = 115°, calculate the following:
(Round your answers to the nearest thousandths place)
a. sin 𝜃
b. cos 𝜃
c. tan 𝜃
7. Given = 55°, calculate the following:
(Round your answers to the nearest thousandths place)
a. sin 𝜃
b. cos 𝜃
c. tan 𝜃
Draft: 3/22/2016
Page 14
8. Using the given triangle, determine:
a. sin 𝜃
39.1m
b. cos 𝜃
c.
tan 𝜃
25m
θ
30m
(Diagram not to scale)
d. Determine 𝜃 to the nearest
tenth of a degree.
9. Using the given triangle, determine:
a. sin 𝜃
60.4mm
17mm
b. cos 𝜃
θ
58mm
c.
tan 𝜃
(Diagram not to scale)
d. Determine 𝜃 to the nearest
tenth of a degree.
10. Using the given triangle, determine
a. sin 𝜃
27m
b. cos 𝜃
θ
81m
c.
tan 𝜃
(Diagram not to scale)
d. Determine 𝜃 to the nearest tenth of
a degree.
Draft: 3/22/2016
Page 15
Appendix H: Day 3 – Pythagorean Theorem and Trig Functions:
Answer Key
PROBLEM
1. Answer:
𝑥 = 53.0𝑐𝑚
Solution:
𝑎2 + 𝑏 2 = 𝑐 2 Where 𝑎 = 45𝑐𝑚 𝑎𝑛𝑑 𝑏 = 28𝑐𝑚
(45𝑐𝑚)2 + (28𝑐𝑚)2 = 𝑥 2
2025𝑐𝑚2 + 784𝑐𝑚2 = 𝑥 2
So 𝑥 2 = 2809𝑐𝑚2 ⟹ 𝑥 = √2809𝑐𝑚2
𝑥 = 53.0𝑐𝑚
2. Answer
𝑥 = 73.4 𝑘𝑚
Solution:
𝑎2 + 𝑏 2 = 𝑐 2 Where 𝑎 = 30𝑘𝑚 𝑎𝑛𝑑 𝑏 = 67𝑘𝑚
(30𝑘𝑚)2 + (67𝑘𝑚)2 = 𝑥 2
900𝑘𝑚2 + 4489𝑘𝑚2 = 𝑥 2
So 𝑥 2 = 5389𝑘𝑚2 ⟹ 𝑥 = √5389𝑘𝑚2
𝑥 = 73.4 𝑘𝑚
3. Answer:
𝑥 = 121.6𝑚𝑚
Solution:
𝑎2 + 𝑏 2 = 𝑐 2 Where 𝑎 = 87𝑚𝑚 𝑎𝑛𝑑 𝑏 = 85𝑚𝑚
(87𝑚𝑚)2 + (85𝑚𝑚)2 = 𝑥 2
7569𝑚𝑚2 + 7225𝑚𝑚2 = 𝑥 2
So 𝑥 2 = 14794𝑚𝑚2 ⟹ 𝑥 = √14794𝑚𝑚2
𝑥 = 121.6mm
4. Answer:
𝑥 = 59.6𝑐𝑚
Solution:
𝑎2 + 𝑏 2 = 𝑐 2 Where 𝑎 = 74𝑐𝑚 𝑎𝑛𝑑 𝑐 = 95𝑐𝑚
(74𝑐𝑚)2 + 𝑥 2 = (95𝑐𝑚)2
𝑥 2 = 9025𝑐𝑚2 − 5476𝑐𝑚2
So 𝑥 2 = 3549𝑐𝑚2 ⟹ 𝑥 = √3549𝑐𝑚2
𝑥 = 59.6𝑐𝑚
5. Answer:
𝑥 = 49.1𝑘𝑚
Solution:
𝑎2 + 𝑏 2 = 𝑐 2 Where 𝑎 =
Draft: 3/22/2016
20𝑘𝑚 𝑎𝑛𝑑 𝑐 = 53𝑘𝑚
Page 16
(20𝑘𝑚)2 + 𝑥 2 = (53𝑘𝑚)2
𝑥 2 = 2809𝑘𝑚2 − 400𝑘𝑚2
So 𝑥 2 = 2409𝑘𝑚2 ⟹ 𝑥 = √2409𝑘𝑚2
𝑥 = 49.1𝑘𝑚
6. Using a calculator the answers are:
a. 0.906
b. -0.423
c. -2.145
7. Using a calculator the answers are:
a. 0.819
b. 0.574
c. 1.428
8. Answer:
𝑂𝑝𝑝𝑜𝑠𝑖𝑡𝑒
25𝑚
𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡
30𝑚
a.
𝑠𝑖𝑛𝜃 = 𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑒𝑠𝑒 = 39.1𝑚 ≈ 0.640
b.
𝑐𝑜𝑠𝜃 = 𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑒𝑠𝑒 = 39.1𝑚 ≈ 0.768
c.
𝑡𝑎𝑛𝜃 = 𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡 = 30𝑚 ≈ 0.833
𝑂𝑝𝑝𝑜𝑠𝑖𝑡𝑒
25𝑚
d. Using any of the answers found in a-c, just use the inverse trig function. This is an
example of using the answer found in a:
𝜃 = 𝑠𝑖𝑛−1 (0.640) = 39.8°
9. Answer:
𝑂𝑝𝑝𝑜𝑠𝑖𝑡𝑒
17𝑚𝑚
𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡
58𝑚𝑚
a.
𝑠𝑖𝑛𝜃 = 𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑒𝑠𝑒 = 60.4𝑚𝑚 ≈ 0.281
b.
𝑐𝑜𝑠𝜃 = 𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑒𝑠𝑒 = 60.4𝑚𝑚 ≈ 0.960
c.
𝑡𝑎𝑛𝜃 = 𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡 = 58𝑚𝑚 ≈ 0.293
𝑂𝑝𝑝𝑜𝑠𝑖𝑡𝑒
17𝑚𝑚
d. Using any of the answers found in a-c, just use the inverse trig function. This is an
example of using the answer found in a:
𝜃 = 𝑠𝑖𝑛−1 (0.281) = 16.3°
10. Answer:
First you must find the Hypotenuse using the Pythagorean Theorem:
𝑎2 + 𝑏 2 = 𝑐 2 ∴ 𝑐 = √(81𝑚)2 + (27𝑚)2 ≈ 85.4𝑚
Now use the Trig definitions:
𝑂𝑝𝑝𝑜𝑠𝑖𝑡𝑒
27𝑚
𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡
81𝑚
a.
𝑠𝑖𝑛𝜃 = 𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑒𝑠𝑒 = 85.4𝑚 ≈ 0.316
b.
𝑐𝑜𝑠𝜃 = 𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑒𝑠𝑒 = 85.4𝑚 ≈ 0.949
c.
𝑡𝑎𝑛𝜃 = 𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡 = 81𝑚 ≈ 0.333
𝑂𝑝𝑝𝑜𝑠𝑖𝑡𝑒
27𝑚
d. Using any of the answers found in a-c, just use the inverse trig function. This is an
example of using the answer found in a:
𝜃 = 𝑠𝑖𝑛−1 (0.316) = 18.4°
Draft: 3/22/2016
Page 17
Appendix I: Day 4 – Inquiry into Refraction
Name _____________________________
Scientists and engineers often conduct experiments that collect a lot of data, that’s the easy part.
What to do with the data is the difficult part! For instance, a very straightforward investigation into
refraction could be done easily by the following:
Take a look at the figure. It is a semi-curricular dish filled with
water. Imagine that a ray of light (like from a laser pointer) is
aimed at the flat side of the dish at the exact center of the dish.
The angle it makes with the dish is called the angle of incidence
and it can be measured. As the ray of light passes into the water
it will be refracted. Because the index of refraction of water is
greater than that of air, the ray of light will be “bent” towards the
normal. This is the angle of refraction and it too can be
measured. The ray will continue in a straight line until it reaches
the curved side of the dish. Because the angle of incidence
when it gets there is 0°, the ray will not be refracted as it exits
the dish. You can record the 2 angle measurements, change the
angle at which the ray of light hits the dish and measure and
record the 2 angles again. You can repeat this process until you
collect a complete set of data.
Normal
Θi
Air
Water
Θr
ΘI = Angle of incidence
Θr = Angle of incidence
Your assignment:
You are to conduct an experiment as described above. After you have collected your data, you
need to analyze the data to determine if there is any relationship between air, water, the incident
angle and the angle of refraction. If a relationship does exist, what is it? If not, suggest an
experiment that you might be able to conduct that will help explain the refraction of light.
Your materials:
1 Laser (either a laser pointer or line level laser)
1 hemi-cylindrical dish filled with water
1 Protractor
Calculator and/or a computer
Pencil and/or pen
Paper
Draft: 3/22/2016
Page 18
SAFETY ALERT!!
As you are well aware, lasers are dangerous if used inappropriately! Therefore:
DO NOT look directly at the laser when it is on
DO NOT point the laser at anyone
Failure to follow the above (as well as all other lab safety policies) may result in immediate
removal from the lab, a grade of “0” for the day and the lab as well as other disciplinary action
suited to the offense.
What needs to be turned in:
You will need to prepare a written report that addresses all of the following:
1. A written description of how you conducted the experiment with an explanation of what
data you collected.
2. A presentation of the data (a data table is highly suggested). Make sure the data is
properly labeled.
3. A written description of how you analyzed your data.
4. A written explanation of what you determined from the data analysis (make sure you
address the questions posed in the “Your assignment” section).
Some tips and suggestions:
 Don’t be afraid to graph your data
 Don’t be afraid to “manipulate” your data
 If in doubt, try something!!
 In science, relationships are often described in the form of a mathematical equation.
Draft: 3/22/2016
Page 19
Appendix J: Day 4 – Inquiry into Refraction: Teacher Notes
This investigation has several objectives for the students. They are as follows:
1.
2.
3.
4.
5.
Conduct a successful experiment
Collect and analyze data
Draw conclusions from the data analysis
Successfully verify Snell’s Law
Clearly communicate their findings.
As the students begin to analyze their data, guide them towards using a spreadsheet program to
graph the data. The initial plots will most likely lead to nothing, so guide them towards looking for
a linear relationship. This is easy to analyze by performing a linear regression. Lead the
students to try to manipulate the data somehow, for example squaring the “y” data and replotting. The correct data manipulation for this activity is to create a plot of the sine of the angle
of incidence versus the sine of the refracted angle. Doing so will result in a straight
line. Performing a linear regression on this plot will lead to a slope of approximately 1.33 which is
the index of refraction for water.
Here is an example data set with the related graph:
Incident Angle
(I)
0
5
10
15
20
25
30
35
40
45
50
55
60
65
70
75
80
Draft: 3/22/2016
Refracted
Angle (R)
0
3.8
7.5
11.3
14.9
18.5
22.1
25.4
28.9
32.1
35.3
38
40.6
42.9
45
46.6
47.8
sin(I)
0.000
0.087
0.174
0.259
0.342
0.423
0.500
0.574
0.643
0.707
0.766
0.819
0.866
0.906
0.940
0.966
0.985
sine(R)
0.000
0.066
0.131
0.196
0.257
0.317
0.376
0.429
0.483
0.531
0.578
0.616
0.651
0.681
0.707
0.727
0.741
Page 20
The mathematical analysis is as follows:
𝑦 = 𝑚∗𝑥+𝑏
𝑇ℎ𝑒 𝑦 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 𝑖𝑠 0 𝑠𝑜 𝑏 = 0
sin(𝜃𝐼 ) = 1.33 sin(𝜃𝑅 )
When you realize that 1.33 is the index of refraction for water, this equation can be generalized
into what is accepted as Snell’s Law:
𝑛𝐼 ∗ sin(𝜃𝐼 ) = 𝑛𝑅 ∗ sin(𝜃𝑅 )
𝑛𝐼 = 𝑖𝑛𝑑𝑒𝑥 𝑜𝑓 𝑟𝑒𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑖𝑛𝑐𝑖𝑑𝑒𝑛𝑡 𝑚𝑒𝑑𝑖𝑢𝑚
𝜃𝐼 = 𝑎𝑛𝑔𝑙𝑒 𝑜𝑓 𝑖𝑛𝑐𝑖𝑑𝑒𝑛𝑐𝑒
𝑛𝑅 = 𝑖𝑛𝑑𝑒𝑥 𝑜𝑓 𝑟𝑒𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑟𝑒𝑓𝑟𝑎𝑐𝑡𝑖𝑣𝑒 𝑚𝑒𝑑𝑖𝑢𝑚
𝜃𝑅 = 𝑎𝑛𝑔𝑙𝑒 𝑜𝑓 𝑟𝑒𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛
Draft: 3/22/2016
Page 21
Appendix K: Day 4 – Inquiry into Refraction Rubric
4
3
2
1
The report is free
from all spelling
and grammatical
errors.
The report has
one spelling
and/or
grammatical
errors.
The report has
two to three
spelling and/or
grammatical
errors.
The report has
four or more
spelling and/or
grammatical
errors.
The description is
detailed, clear,
complete and
easy to
understand.
The description is
clear, complete
and easy to
understand.
The description is
not clear and not
easy to
understand.
The report is
seriously lacking
in any substance.
The data is
presented in a
detailed, clear,
complete and
easy to
understand
format.
The data is
presented in a
clear, complete
and easy to
understand
format.
The data is not
presented in a
clear and easy to
understand
format.
The data is not
presented in any
understandable
format.
Description of
the Data
Analysis
The description is
detailed, clear,
complete and
easy to
understand.
The description is
clear, complete
and easy to
understand
The description is
not clear and not
easy to
understand.
The report is
seriously lacking
in any substance.
Explanation
of what was
determined
The description is
detailed, clear,
complete and
easy to
understand.
The description is
clear, complete
and easy to
understand
The description is
not clear and not
easy to
understand.
The report is
seriously lacking
in any substance
Spelling and
grammar
Description of
experiment
Data
Presentation
Draft: 3/22/2016
Page 22
Appendix L: Day 6 – Engineering Design Challenge
Fantastic news! Your communications company has been selected to demonstrate how it is
possible to send information from a ground-based transmitter to a receiver that is located
underwater!
As you know your company’s major focus is transmitting information by using light, much like
today's cable TV companies. The big difference is your company uses lasers rather than a cable.
The challenge that your company is facing is as follows:
Your company is to engineer a prototype communication system to transmit a land based signal
using the principles of reflection and refraction that will be redirected from a tower to
communicate with an object located in a body of water. Because your company can send
information using lasers, all you need to do is demonstrate how you can “hit” an object
submerged in water.
A successful prototype will be able to hit two separate objects that will be submerged in an
aquarium. You will be given a range of 20 cm where your laser must enter the water. You will be
given a time limit of 10 minutes per object to set up your communications system and only 2
chances to “hit” each object.
The materials that will be supplied:
1 Laser
1 Flat mirror
1 Aquarium filled with water
2 Protractors
1 Meter stick
1 Ruler
1 Object
You are free to use any materials (within reason) to construct your communications system as
long as you receive prior approval from me!
Draft: 3/22/2016
Page 23
Appendix M: Day 6 – Engineering Design Challenge Rubric
Company Member’s Names
___________________________
__________________________
___________________________
__________________________
This rubric will be used for each object. Therefore there are 16 points possible.
4
Time limit
Execution
3
2
Completed
within the time
limit
Hit the object
on the first try
Ran over the
time limit
Hit the object
on the second
try
Either attempt
was within 10
cm of the
object
Score for Object 1
______________________
Score for Object 2
______________________
Total Score
______________________
Draft: 3/22/2016
1
Either attempt
was within 20
cm of the
object
Page 24
Appendix N: Day 6 – Daily Performance Rubric
CATEGORY
4
Safety
The student
follows all safety
policies.
Preparedness
Attitude
Focus on the
task
Working with
Others
Draft: 3/22/2016
3
2
1
The student fails
to follow a safety
policy (This
results in a “0”
for this category)
Brought all
needed
materials to
class
Failed to bring
one item needed
for the day’s
work.
Failed to bring
more than one
item needed for
the day’s work.
Failed to bring
anything for
class.
Never is publicly
critical of the
project or the
work of others.
Always has a
positive attitude
about the
task(s).
Observed one
time to be
publicly critical of
the project or the
work of others.
Often has a
positive attitude
about the
task(s).
Observed twice
to be publicly
critical of the
project or the
work of other
members of the
group. Usually
has a positive
attitude about
the task(s).
Observed more
than twice to be
publicly critical of
the project or the
work of other
members of the
group. Often has
a negative
attitude about
the task(s).
Consistently
stays focused on
the task and
what needs to be
done. Very selfdirected.
Observed to be
off task once.
Other group
members can
count on this
person.
Observed to be
off task twice.
Other group
members must
sometimes nag,
prod, and remind
to keep this
person on-task.
Observed to be
off task more
than twice. Lets
others do the
work.
Almost always
listens to, shares
with, and
supports the
efforts of others.
Tries to keep
people working
well together.
Observe 80% of
the time listens
to, shares, with,
and supports the
efforts of others.
Does not cause
"waves" in the
group.
Observe less
than 80% of the
time listens to,
shares with, and
supports the
efforts of others,
but sometimes is
not a good team
member.
Does not listen
to, share with,
and support the
efforts of others.
Often is not a
good team
player.
Page 25
Appendix I: Day 6 – Engineering Design Process
Ask:
What is the problem?
What have others done?
What are the constraints?
Think:
What could be some solutions?
Brainstorm ideas; choose the best ones.
Plan:
Draw a diagram.
Make a list of materials, you will need.
Test:
Follow your plan and create it.
Test your solution to the problem.
Improve:
Make the design better.
Test it, again.
Draft: 3/22/2016
Page 26
Appendix P: Additional Teacher Resources
Demonstration of refraction and reflection (Note: it is very long, you might want to only
use a couple minutes of it)
http://www.youtube.com/watch?v=2kBOqfS0nmE
Pictures of Reflection and Refraction
http://www.flickr.com/photos/physicsclassroom/galleries/72157625102649965/
Laser Communication Video links:
http://www.youtube.com/watch?v=HKRPfa66_po
http://www.youtube.com/watch?v=mLo8h3tlyvo
http://www.youtube.com/watch?v=J-kZpuNA57c&list=PLDDC2C4370524AC28
http://www.youtube.com/watch?v=Vi5jnExnjxA
NASA article about satellite laser communication
http://www.gizmag.com/nasa-laser-communications-relay-demonstration/19946/
The Physics Classroom
Excellent site for all things related to physics!!
http://www.physicsclassroom.com/Class/refrn/u14l2b.cfm
PHet Sim
Excellent site that has a lot of simulations
http://phet.colorado.edu/en/simulation/bending-light
Science Kit
This is where you can obtain the hemi-cylindrical dishes
http://sciencekit.com/semicircular-refraction-cells/p/IG0035616/
Laser Safety InformationLaser Institute of America
http://www.lia.org/subscriptions/safety_bulletin/laser_safety_information
Penn State Laser Safety
http://www.ehrs.upenn.edu/programs/laser/lasermanual/
OSHA
http://www.osha.gov/SLTC/laserhazards/
The Engineering Design Process
This is a great site to help you understand what the Engineering Design Process is.
http://www.nasa.gov/audience/foreducators/plantgrowth/reference/Eng_Design_512.html
Draft: 3/22/2016
Page 27
Appendix Q: Technical Brief - Refraction and Reflection of Light
When light travels through a uniform material, such as air or glass, it travels in a straight line. The
direction of the light can be described by a ray oriented along the line and pointed in the direction
of energy flow. When the ray encounters an interface between two materials, however, the
direction of the light can change. In fact, the energy can be split between two rays, one reflected
from the interface and one refracted into the new material as illustrated in Figure 1. Both the
amount of energy split between these two rays and the directions of the reflected and refracted
rays depend on a material property known as the index of refraction. In this technical brief, we
explore how and why the ray directions change at the material interface.
Figure 1. Refraction and reflection at an interface between two materials.
Light reflection and refraction arise because of the wave properties of light. Although it is
common to consider the propagation, or movement, of light in terms of rays, it is important to
understand that it is actually an electromagnetic wave, as depicted in Figure 2, with electric and
magnetic fields that oscillate in both space and time. The oscillation is time is often
characterized a frequency, or oscillation rate, and the oscillation in space can be characterized by
a wavelength, or distance between the peaks in the electric field as shown in the figure. The
speed at which the wave propagates is the product of the frequency and wavelength. If we use
the Greek letter ν (nu) to represent the frequency and λ (lambda) to represent the wavelength,
then the speed of light c = λν.
Draft: 3/22/2016
Page 28
Figure 2. Depiction of light as an electromagnetic wave.
When light is traveling in free space where there are no molecules or atoms to interact with, the
speed of light is 3 x 108 m/s. However, when it travels in any other material, the interaction with
the molecules and atoms of which the material is composed causes it to slow down. This is
characterized by a material property know as the index of refraction, which we will designate by n.
The index of refraction is one for free space; it is approximately 1.0 for air, 1.3 for water, and 1.41.6 for glass. Consider a light beam encountering a material interface straight-on as shown in
Figure 3. To illustrate the wave property, we depict a number of wavefronts marching along in
the propagation direction, each corresponding to a peak in the electric field (much like peaks of a
wave in water) and separated by the wavelength. Inside the material, both the speed and the
wavelength become smaller by the factor 1/n while the frequency stays the same. In this case,
the direction of the light does not change.
Figure 3. Decrease in speed and wavelength of light in a material.
Draft: 3/22/2016
Page 29
When light encounters and interface at an angle, things become a bit more interesting. Using
Figure 4 as a guide, we first consider the direction of the refracted light at the interface between
two materials, one with an index of refraction denoted by n1 and another denoted by n2. The
subscript in each merely indicates the material. We also use the variable θ1 to represent the
angle at which the incoming ray encounters the interface. It is measured from a perpendicular
line to the surface indicated by the dashed line, such that an angle of zero corresponds to the
straight-on situation described in Figure 3. What we want to figure out is the angle θ2, measured
in the same way, of the refracted ray in the second material. According to the wave property of
light, all the wavefronts in both materials must line up at the interface, and this ties together the
ray directions in both materials. Examining one period of the waves along the interface, as
shown by the magnified view on the right side of Figure 4, basic trigonometry requires that
𝜆/𝑛1
𝜆/𝑛2
𝑠𝑖𝑛𝜃1 =
and 𝑠𝑖𝑛𝜃2 =
.
𝑑
𝑑
With a little algebra (solving both for d, setting them equal to each other, and rearranging a little),
we see that
𝑛1 𝑠𝑖𝑛𝜃1 = 𝑛2 𝑠𝑖𝑛𝜃2 .
This is known as Snell’s Law. If one knows the index of refraction of both materials, then the
refracted ray direction θ2 can be solved for any incident ray direction θ1.
Figure 4. Wavefront matching for refraction at the interface between two materials.
The reflected ray direction can be found in the same way, but the situation is actually much
simpler as shown in Figure 5. In this case, the wavelength for the reflected ray is the same as the
incident ray because it is still in the same material. Therefore, the wavefronts will only match up
at the interface if the reflection is symmetric about the line perpendicular to the interface. That is,
𝜃1 = 𝜃2 .
This is known as the Law of Reflection.
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Figure 5. Wavefront matching for reflection at the interface between two materials.
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