For problems, the correct equation(s) is stated. The unknown is algebraically isolated. The given values are substituted in. A reasonable amount of correct steps or rationale are shown (even for...

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For problems, the correct equation(s) is stated. The unknown is algebraically isolated. The given values are substituted in. A reasonable amount of correct steps or rationale are shown (even for multiple choice). The correct answer is given, with appropriate units and direction if needed. The answer is rounded as required. All parts of the question are answered. Words and/or sketches are used to illustrate your point if appropriate. Work is laid out in a neat, readable, logical flow. For explanations, answers are as clear, specific and succinct as possible. They could be in words or diagrams or both. All parts of the question are answered. The work shows the full extent of your understanding using the relevant concepts accurately.


Microsoft Word - Set 3.docx conted.ucalgary.ca Physics 2 Set 3 Lesson 5. Light, Reflection, Mirrors. 1. As the frequency of waves increases, the wavelength __________ . The range of wavelengths for visible light is from __________nm to __________nm. Which type of wave would penetrate the human body more easily, X-rays or gamma rays? Circle correct response. 2. Calculate the quantity indicated for each of the following electromagnetic waves: (a) The frequency of a 1.8 cm microwave. (1.7 x 1010 Hz) (b) The wavelength of a 3.2 x 1010 Hz radar signal (a radio wave). (9.4 mm) 3. In communicating with a space station, radio signals traveling at the speed of light must travel a distance of 8.7×109 m each way. Calculate the total time it takes for a radio signal to travel to the station and back. (58 s) 4. Our nearest neighbor among the stars (aside from our own Sun) is Proxima Centauri, which is located 4.3 light years away. Find the: (a) Distance to Proxima Centauri in kilometers. (4.07×1013km) (b) Time in years it would take our fastest currently imagined spacecraft (maximum speed of 3.0×105 m/s) to reach Proxima Centauri. (~4300 years) 2 5. Draw a ray diagram and image for each of the arrows (objects) in front of the curved mirrors. Use a ruler, please. Use 2-3 rays. Describe each image as real or virtual. Describe each device as concave or convex, diverging or converging. (a) circle one in each pair: real / virtual concave / convex diverging / converging (b) circle one in each pair: real / virtual concave / convex diverging / converging F C F C Mirrored this side Mirrored this side 3 (c) circle one in each pair: real / virtual concave / convex diverging / converging 6. A 6.0 cm tall object is placed 40 cm in front of a concave mirror with a radius of curvature of 60 cm. Draw a small ray diagram and the image (no ruler needed), and calculate the: (a) Distance from the mirror to the image (1.2 m) (b) Height of the image (-18 cm) (c) Describe the image formed as virtual or real, and inverted or upright Lesson 6. Refraction, Lenses. 7. If the speed of light in a specific plastic is 2.0×108 m/s, calculate the index of refraction in this plastic. (1.5) 8. If the index of refraction for crown glass is 1.52, determine the speed of light in crown glass. (1.97×108 m/s) C ’ F ’ Mirrored this side 4 9. A beam of light travelling in air strikes the surface of water (n = 1.33) with an incident angle of 60°. Draw a small sketch with labelled angles and media (no ruler needed), and find the angle of refraction as the beam travels into the water. (41°) 10. The index of refraction of diamond is 2.42. Use the diagram below and knowledge of the phenomenon of total internal reflection (TIR) to explain why diamonds that are cut correctly are much more sparkly than a piece of glass of the same cut. What happens to incoming rays of light that is different from glass? Source: https://www.physicsclassroom.com/class/refrn/Lesson-3/The-Critical-Angle 11. Fill in Table 1. Table 1. Curved Mirrors and Thin Lenses Facts Diverging Devices Converging Devices Mirror type Con______ Which type(s) of image? Circle: real and/or virtual Con______ Which type(s) of image? Circle: real and/or virtual Lens type Con______ Which type(s) of image? Circle: real and/or virtual Con______ Which type(s) of image? Circle: real and/or virtual Focal length f Circle one: + - Circle one: + - Distance from object to device do Circle one: + - Circle one: + - Height of object ho Circle one: + - Circle one: + - 12. Fill in Table 2. Table 2. Real and Virtual Image Facts Virtual image Real image Distance from image to device di Circle one: + - Circle one: + - Height of image hi Circle one: + - Circle one: + - 5 13. Draw a ray diagram and image for each of the following lenses. Use a ruler please. Use 2- 3 rays. Describe each image as real or virtual. (a) circle one in each pair: real / virtual concave / convex diverging / converging (b) circle one in each pair: real / virtual concave / convex diverging / converging (c) circle one in each pair: real / virtual concave / convex diverging / converging 6 14. A lamp 10 cm tall is placed 60 cm in front of a concave lens with a focal length of 20 cm. Draw a ray diagram (no ruler needed) and the image, and determine the: (a) Distance from the image to the center of the lens (-15 cm: yes this is correct) (b) Height of the image (2.5 cm) Refraction Activity. Let's explore refraction. Navigate to the applet: https://phet.colorado.edu/en/simulation/bending- light?rel=0&showinfo=0&iv_load_policy=3&cc_load_policy=1 If your computer does not allow it to run, select HTML5 if it is an option. If no luck, ask me for troubleshooting – a last resort is using simulated data I can provide. Alternately, you may wish to do this activity with real objects. This is very cool! You will need a laser, a straight walled container for water with an open top and a clear bottom (such as a small aquarium or rectangular vase), and a protractor. You can use any depth of water. To make the laser path more visible in the water, try adding a tiny pinch of baby powder or a drop of milk to the water. If you are using the applet, enter the Intro tab. Set it up as shown with water in the top, air in the bottom. This is a top view looking downward. Place the protractor with its centre on the boundary between water and air. Aim the laser through 0 degrees. This is a zero degree angle of incidence. (NOT 90 degrees!!!) Hypotheses: 1. If a laser is shone into water at increasing incident angles, it will exit into air at a refracted angle that is larger than the incident angle, and according to Snell’s Law. 2. When the incident angle reaches the critical angle, the ray will become totally internally reflected with no light escaping the water. 7 Source: http://labman.phys.utk.edu/phys222core/modules/m7/internal-reflection.html Variables: Manipulated (independent variable): qi angle of incidence Responding (dependent variables): qr angle of refraction qR angle of internal reflection Controlled variables: wavelength of light, air / water temperature and density, measuring technique Procedure: Turn on the laser. Play around for different ??, observing ??, ??, and their intensity. Now set your angle of incidence to the first value in Table 3. For each input of the laser qi, measure and record the output qr (refraction) and qR (internal reflection) using the protractor. ?? ?? ?? 8 Table 3. Refraction / Total Internal Reflection, Medium 1 n=1.33, Medium 2 n=1.00 Input Observations, Quantitative: Observations, Qualitative: Analysis: (should be approx. 1.33) qi angle of incidence qr angle of refraction qR angle of reflection At which qi did you start to see any internal reflection? What happened to the internally reflected beam as you increased the angle of incidence? What happened to the refracted beam as you increased the angle of incidence? At which approximate angle did you see total internal reflection? How would you describe the refracted rays for incident angles greater than the critical angle? Sinqr / Sinqi (calculated to 3 decimals) 0 0 n/a n/a 10 20
Answered Same DayOct 21, 2021

Answer To: For problems, the correct equation(s) is stated. The unknown is algebraically isolated. The given...

Ravindra Kumar answered on Oct 22 2021
121 Votes
Solution 1
As the frequency of waves increases, the wavelength __decreases___. The range of
wavelengths for visible light is from ___380 ____ nm to ___700_______nm. Which type of
wave would penetrate the human body more easily, X-rays or gamma rays? Ci
rcle correct
response.
Reason:
We know that the relation between the frequency and the wavelength of the wave

But for the case of the monochromatic light source, if the wavelength of the light is getting change then it will not affect the frequency of the light. In that case the frequency of the light remains unchanged.
The penetration power of gamma rays higher than the penetration power of x-rays. Because we know that the frequency of the gamma rays is but the frequency of the x-rays is , hence we can say that the energy of the gamma rays is higher than the energy of the x-rays.
Solution 2
a)
Given,
Wavelength of the microwave
Speed of light
Frequency of the wave
b)
Given,
Frequency
Speed of light
Solution 3
Given,
Distance
Speed of light
Time (t) =?
Here t is the time taken by the radio wave to reach the station.
So, radio wave goes to the station and come back then total distance need to travel will be times so the time taken by the radio signal will be 2t. hence the required time will be 58 sec.
Solution 4
Given,
The distance between the Proxima Centauri
a)
We know that the 1 light year
So, the distance between the Proxima Centauri and earth is
b)
Speed of the spacecraft (v)
Time = distance / speed
Hence, the required time in hour

Now, converting the hour into the year.
Hence, total required time in year
Solution 5
From the given diagram, we have sketched the ray diagram, which is given below -
a)
Here,
Image is real, diminished and inverted.
Mirror is concave mirror
And it is conversing the ray of light.
b)
From the above ray diagram which can understand that when the object is at the centre of the concave mirror then the image will be real, inverted and equal to the size of the object.
Hence, image is real, concave and converging.
c)
From the ray diagram, we can understand that the image is virtual and diminished. Nd the mirror is convex mirror and rays will be diverging.
· Virtual
· Convex mirror
· Virtual Image
Solution 6
Given,
Height of the object
Distance between the object and the concave mirror
Radius of the curvature of the concave mirror
Hence, focal length of the mirror
So,
a)
Let the distance from the mirror to the image be v,
b)
Height of the image for the concave...
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