Dayton Regional Stem Centerdaytonregionalstemcenter.org/wp-content/uploads/2013/06/... · Web viewThe index of refraction is one for free space; it is approximately 1.0 for air, 1.3

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Printable ResourcesBending 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

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Appendix A: Day 1 – Pre/Post TestName _____________________________

ProblemsShow all of work.

1. Solve the following equation for the variable b.a∗tanθ=b∗tanθ

2. Solve for the variable a in the following equation when c = 14 and b = 10a2+b2=c2

3. Using the right triangle below, find the value of x to the nearest tenth.

4. Using the right triangle below, give the correct ratios for the sine, cosine and tangent of angle C.

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(Diagram not to scale)

28 cm

45 cm

x


12

C

13

5. Using the right triangle below, calculate angle c to the nearest whole degree.

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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6

C

8

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∗tanθ=b∗tanθ

b=a∗tan θtanθ

Points Rationale

4 The student correctly calculates the answer.3 The student makes one error in calculating the answer.2 The student makes more than one error in calculating the answer.1 The student attempts to solve the problem but is unsuccessful.

2. a2+b2=c2 ∴ a=√c2−b2

a=√142−102=9.798

Points Rationale


3. Answer:x=53.0 cm

Solution:a2+b2=c2 Where a=45cm∧b=28cm(45cm)2+(28cm)2=x2

2025 cm2+784 cm2=x2

So x2=2809 cm2⟹ x=√2809cm2

x=53.0 cm

Points Rationale


4. Answer:

sin C=1213 cosC= 5

13 tanC=125

Points Rationale

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4 The student correctly states all three ratios3 The student makes one error in stating the ratios2 The student makes more than one error in stating the ratios1 The student attempts to solve the problem but is unsuccessful.

5. Answer:

C=sin−1( 68 )=49°

Points Rationale

4 The student correctly calculates the angle.3 The student makes one error in calculating the angle.2 The student makes more than one error in calculating the angle.1 The student attempts to solve the problem but is unsuccessful.

6. Answer:The angle of refraction is calculated using Snell’s Law:

sin θ I∗n I=sinθR∗nR

θR=sin−1[ sin (50 °)1.33 ]=35 °

Points Rationale

4 The student correctly calculates the angle of refraction by using Snell’s Law

3 The student makes one error in calculating the angle of refraction by using Snell’s Law

2 The student makes more than one error in calculating the angle of refraction by using Snell’s Law

1 The student attempts to solve the problem BUT with out using Snell’s Law

7. AnswerUsing 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 sinθ I=nR sin θR

nR=n I sinθ I

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.

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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 Rationale

4 The student correctly calculates the index of refraction AND states that it is NOT enough to conclude the suspect was at the scene.

3 The student correctly calculates the index of refraction BUT states that it IS enough to conclude the suspect was at the scene.

2 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.

1 The student made an attempt but is incorrect.

SHORT ANSWER

8. AnswerAnswers may vary. Applications within the following groups could be used: industrial, environmental, communications, research, medical.

Points Rationale

4 Student listed four applications.3 Student listed three applications.2 Student listed two applications.1 Student listed one application.0 Student did not list any applications

9. Answer:Point

s Rationale

4Student 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)

3 Student omits 1 step from the process2 Student omits 2-3 steps from the process1 Student omits 4-5 steps from the process0 Student cannot name any steps of the process

ESSAY

10. AnswerAnswers will vary. An exemplar response is:Light travels at different speeds in different mediums. In a vacuum, light travels at 3x108 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 Rationale

4 The student describes how light travels at Draft: 9/11/2023 Page 7

different speeds in different mediums AND uses the term “refracted” or “refraction” correctly.

3 The students describes how light travels at different speeds in different mediums BUT uses the term “bend” or “bent” instead of “refracted” or “refraction”.

2 The student uses terms such as “medium” and “refracted” but incorrectly.1 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. Reflection2. Refraction3. Incidence4. Wavefront5. Laser6. Index of refraction7. Speed8. Velocity9. 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 medium5. 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(θ¿¿i)=nr∗sin(θ¿¿r)¿¿¿¿Where θi ("theta i") = angle of incidence θr ("theta r") = angle of refractionni = index of refraction of the incident mediumnr = 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:

a2 + b2= c2

By definition, the Pythagorean Theorem is:

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:

Solution:

a2+b2=c2 Where a=27cm∧b=73cm(27cm)2+(73cm)2=x2

729cm2+5329 cm2=x2

So x2=6058 cm2⟹ x=√6058 cm2

x=77.8 cm

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:

Solution:

a2+b2=c2 Where a=21cm∧c=74 cm(21cm)2+x2=(74 cm)2

x2=5476 cm2−441cm2

So x2=5035 cm2⟹ x=√5035 cm2

x=71.0 cm

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x

27 cm

73 cm


74 cm

21 cm

x


b

a

c

θ

39.1m

30m

25m


θ

98.5m

98m

10m


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 sidehypotenuse

=bc

cos B=adjacent sidehypotenuse

=ac

tan B=opposite sideadjacent side

=ba

sin A¿ opposit e sidehypotenuse

=ac

cos A=adjacent sidehypotenuse

=bc

tan A= opposite sideadjacent side

=ab

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 θb. cosθc. tanθ

a. sin θ= oppositehypotenuse

= 25m39.1 m

≈ 0.640

b. cosθ= adjacenthypotenuse

= 30m39.1m

≈ 0.768

c. tanθ= oppositeadjacent

=25m30m

≈ 0.833

Using the triangle given,determine the following:



= 10m98.5 m

≈ 0.102


= 98m98.5 m

≈ 0.995


=10m98m

≈ 0.102

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A

a

c

B

b

θ90m

26m


It might be necessary to employ the Pythagorean Theorem before you can find all the trigonometric relations. Here is an example:

Using the triangle given,determine the following:


First you must find the Hypotenuse using the Pythagorean Theorem:

a2+b2=c2 ∴c=√(90 m)2+(26 m)2≈ 93.7 m

Now continue using the Trig definitions:


= 26 m93.7 m

≈ 0.278


= 90m93.7 m

≈ 0.961


=26m90m

≈ 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.

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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:



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85mm

87mm

x


28 cm

45 cm

x


67km

30km

x



6. Given θ = 115°, calculate the following: (Round your answers to the nearest thousandths place)

a. sin θ

b. cosθ

c. tanθ

7. Given = 55°, calculate the following:(Round your answers to the nearest thousandths place)

a. sin θ

b. cosθ

c. tanθ

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x

74cm

95cm


x

20km

53km

8. Using the given triangle, determine:

a. sin θ

b. cosθ

c. tanθ

d. Determine θ to the nearest tenth of a degree.

9. Using the given triangle, determine:

a. sin θ

b. cosθ

c. tanθ


10. Using the given triangle, determine

a. sin θ

b. cosθ

c. tanθ


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θ

39.1m

30m

25m


θ

60.4mm

58mm

17mm


θ81m

27m


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Appendix H: Day 3 – Pythagorean Theorem and Trig Functions: Answer Key

PROBLEM

1. Answer:x=53.0 cm

Solution:a2+b2=c2 Where a=45cm∧b=28cm(45cm)2+(28cm)2=x2

2025 cm2+784 cm2=x2

So x2=2809 cm2⟹ x=√2809cm2

x=53.0 cm

2. Answerx=73.4 km

Solution:a2+b2=c2 Where a=30km∧b=67km(30km)2+(67km)2=x2

900 km2+4489 km2=x2

So x2=5389 km2⟹ x=√5389 km2

x=73.4 km

3. Answer:x=121.6 mm

Solution:a2+b2=c2 Where a=87mm∧b=85mm(87mm)2+(85 mm)2=x2

7569 mm2+7225 mm2=x2

So x2=14794 mm2⟹ x=√14794 mm2

x=121.6mm

4. Answer:x=59.6 cm

Solution:a2+b2=c2 Where a=74cm∧c=95cm(74 cm)2+ x2=(95 cm)2

x2=9025 cm2−5476 cm2

So x2=3549 cm2⟹ x=√3549 cm2

x=59.6 cm

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5. Answer:x=49.1km

Solution:a2+b2=c2 Where a=20km∧c=53km(20km)2+x2=(53 km)2

x2=2809km2−400km2

So x2=2409 km2⟹ x=√2409km2

x=49.1km

6. Using a calculator the answers are:a. 0.906b. -0.423c. -2.145

7. Using a calculator the answers are:a. 0.819b. 0.574c. 1.428

8. Answer:

a. sinθ= OppositeHypotenuese

= 25 m39.1m

≈ 0.640

b. cosθ= AdjacentHypotenuese

= 30m39.1m

≈ 0.768

c. tanθ=O ppositeAdjacent

=25m30m

≈ 0.833

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:

a. sinθ= OppositeHypotenuese

= 17mm60.4mm

≈ 0.281


= 58mm60.4 mm

≈ 0.960

c. tanθ=OppositeAdjacent

=17mm58mm

≈ 0.293


10. Answer:First you must find the Hypotenuse using the Pythagorean Theorem:a2+b2=c2 ∴c=√(81m)2+(27 m)2≈ 85.4 mNow use the Trig definitions:

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a. sinθ= Oppo siteHypotenuese

= 27 m85.4m

≈ 0.316


= 81m85.4 m

≈ 0.949

c. tanθ=OppositeAdjacent

=27m81m

≈ 0.333


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.

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 waterDraft: 9/11/2023 Page 20

ΘI = Angle of incidenceΘr = Angle of incidence

Normal

Water

Air

Θr

Θi

1 ProtractorCalculator and/or a computerPencil and/or penPaper

Draft: 9/11/2023 Page 21

SAFETY ALERT!!As you are well aware, lasers are dangerous if used inappropriately! Therefore:

DO NOT look directly at the laser when it is onDO 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.

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Appendix J: Day 4 – Inquiry into Refraction: Teacher Notes

This investigation has several objectives for the students. They are as follows:

1. Conduct a successful experiment2. Collect and analyze data3. Draw conclusions from the data analysis4. Successfully verify Snell’s Law5. 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 re-plotting. 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)

Refracted Angle (R) sin(I) sine(R)

0 0 0.000 0.0005 3.8 0.087 0.066

10 7.5 0.174 0.13115 11.3 0.259 0.19620 14.9 0.342 0.25725 18.5 0.423 0.31730 22.1 0.500 0.37635 25.4 0.574 0.42940 28.9 0.643 0.48345 32.1 0.707 0.53150 35.3 0.766 0.57855 38 0.819 0.61660 40.6 0.866 0.65165 42.9 0.906 0.68170 45 0.940 0.70775 46.6 0.966 0.72780 47.8 0.985 0.741

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The mathematical analysis is as follows:y=m∗x+bThe y intercept is 0 sob=0

sin (θ I )=1.33sin (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 (θ I )=nR∗sin (θR )n I=index of refraction of the incident medium

θ I=angleof incidencenR=index of refraction of the refractive mediumθR=angle of refraction

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Appendix K: Day 4 – Inquiry into Refraction Rubric

4 3 2 1

Spelling and grammar

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.

Description of experiment

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.

Data Presentation

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 clear, complete and easy to understand


The report is seriously lacking in any substance.

Explanation of what was determined


The description is clear, complete and easy to understand


The report is seriously lacking in any substance

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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 Laser1 Flat mirror1 Aquarium filled with water2 Protractors1 Meter stick1 Ruler1 Object

You are free to use any materials (within reason) to construct your communications system as long as you receive prior approval from me!

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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 3 2 1

Time limitCompleted within the time limit

Ran over the time limit

ExecutionHit the object on the first try

Hit the object on the second try

Either attempt was within 10 cm of the object

Either attempt was within 20 cm of the object

Score for Object 1 ______________________

Score for Object 2 ______________________

Total Score ______________________

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Appendix N: Day 6 – Daily Performance Rubric

CATEGORY 4 3 2 1

Safety

The student follows all safety policies.

The student fails to follow a safety policy (This results in a “0” for this category)

PreparednessBrought 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.

Attitude

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).

Focus on the task

Consistently stays focused on the task and what needs to be done. Very self-directed.

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.

Working with Others

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.

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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.

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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 Refractionhttp://www.flickr.com/photos/physicsclassroom/galleries/72157625102649965/

Laser Communication Video links:http://www.youtube.com/watch?v=HKRPfa66_pohttp://www.youtube.com/watch?v=mLo8h3tlyvohttp://www.youtube.com/watch?v=J-kZpuNA57c&list=PLDDC2C4370524AC28http://www.youtube.com/watch?v=Vi5jnExnjxA

NASA article about satellite laser communicationhttp://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 SimExcellent site that has a lot of simulationshttp://phet.colorado.edu/en/simulation/bending-light

Science KitThis is where you can obtain the hemi-cylindrical disheshttp://sciencekit.com/semicircular-refraction-cells/p/IG0035616/

Laser Safety Information-Laser Institute of America http://www.lia.org/subscriptions/safety_bulletin/laser_safety_informationPenn State Laser Safetyh ttp://www.ehrs.upenn.edu/programs/laser/lasermanual/ OSHAhttp://www.osha.gov/SLTC/laserhazards/

The Engineering Design ProcessThis is a great site to help you understand what the Engineering Design Process is.http://www.nasa.gov/audience/foreducators/plantgrowth/reference/Eng_Design_5-12.html

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http://www.nasa.gov/audience/foreducators/plantgrowth/reference/Eng_Design_5-12.html

http://www.nasa.gov/audience/foreducators/plantgrowth/reference/Eng_Design_5-12.html

http://www.osha.gov/SLTC/laserhazards/

http://www.ehrs.upenn.edu/programs/laser/lasermanual/

http://www.lia.org/subscriptions/safety_bulletin/laser_safety_information

http://sciencekit.com/semicircular-refraction-cells/p/IG0035616/

http://phet.colorado.edu/en/simulation/bending-light

http://www.physicsclassroom.com/Class/refrn/u14l2b.cfm

http://www.gizmag.com/nasa-laser-communications-relay-demonstration/19946/

http://www.youtube.com/watch?v=Vi5jnExnjxA

http://www.youtube.com/watch?v=J-kZpuNA57c&list=PLDDC2C4370524AC28

http://www.youtube.com/watch?v=mLo8h3tlyvo

http://www.youtube.com/watch?v=HKRPfa66_po

http://www.flickr.com/photos/physicsclassroom/galleries/72157625102649965/

http://www.youtube.com/watch?v=2kBOqfS0nmE

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.

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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 1.4-1.6 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.

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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

sin θ1=λ /n1

d and sin θ2=

λ /n2

d .

With a little algebra (solving both for d, setting them equal to each other, and rearranging a little), we see that

n1 sin θ1=n2sin θ2 .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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Documents

Dayton Regional Stem Centerdaytonregionalstemcenter.org/wp-content/uploads/2013/06/... · Web viewThe index of refraction is one for free space; it is approximately 1.0 for air, 1.3