Ray Tracing (Lenses) Gizmo Answer Key

Name: Ari Rosario Date: 07-10-2022 Student Exploration: Ray Tracing (Lenses) Directions: Follow the instructions to go through the simulation. Respond to the questions and prompts in the orange boxes. Vocabulary: concave lens, convex lens, focal point, image, magnification, real image, refraction, virtual image Prior Knowledge Questions (Do these BEFORE using the Gizmo.) Agnes is trapped on a desert island with nothing but a magnifying glass. She wants to use the glass to focus sunlight and start a fire. She holds the glass above some dry grass as shown at right. 1. On the diagram, draw the path the Sun rays will likely take from the magnifying glass to the grass. 2. A magnifying glass is an example of a convex lens—a lens that curves outward on both sides. Why is a convex lens useful for starting fires? Because the lens can focus all of the light on a single place, it causes that point to heat up and possibly catch fire. Gizmo Warm-up The Ray Tracing (Lenses) Gizmo shows light rays passing through a lens. The light rays are bent by refraction as they pass through the lens and form a focused image to the right of the lens. To begin, turn on the Colorize lines checkbox. Under Show lines, turn off the Central line and the Line through focal point so that only the Parallel line is showing. 1. The blue dots in front of and behind the lens are the focal points of the lens. Move the candle on the left back and forth and up and down. What is always true about the light ray that emerges from the right side of the lens? The focal point is always the route it takes. 2. Turn off the Parallel line and turn on the Line through focal point. Move the candle. What do you notice about this line? It always runs perpendicular to the main axis. Activity A: Get the Gizmo ready: Real and virtual images ● Turn on the Parallel line, Central line, and Line through focal point. ● Move the candle to -24 on the central axis, with the focal point at -12. Introduction: A convex lens is called a “converging lens” because it focuses light rays into a point. A real image is formed where the light rays emitted from a point converge on the other side of the lens. If you placed a sheet of paper at the image, a focused image would be projected onto the paper. Question: How do lenses create images? 1. Observe: In its current configuration, the distance from the candle to the focal point is 12 units and the distance from the focal point to the lens is also 12 units. A. What do you notice about the orientation of the candle’s image on the right side of the lens? It is flipped. B. What do you notice about the size of the image? The picture is a little bit smaller. 2. Investigate: Complete each action listed in the table below, and describe how that action affects the image to the right of the lens. Return the candle and focal point to their original positions (-24 for the candle, -12 for the focal point) after each action. Action Effect on image Move the candle to the left. Image gets smaller and moves toward the lens. Move the candle to the right. Image gets bigger and moves away from the lens. Move the left focal point to the left. Image gets bigger and moves away from the lens. Image gets smaller and moves toward the lens. Move the left focal point to the right. 3. Analyze: How is the image size related to the distance between the candle and the focal point? The image shrinks and travels closer to the lens as the distance between the candle and the focal point grows, and the opposite is also true. 4. Explore: Move the candle to -12 and the focal point to -24. Shorten the candle to see the whole image. What do you notice when the candle is between the focal point and the lens? The image is taller and further distant from the lens than the candle, and it is also virtual (the picture is on the same side of the lens as the object). The image you see is called a virtual image because no actual light rays are focused there. The dotted lines represent locations from which light appears to be coming to an observer located left of the lens. An observer would perceive a magnified virtual image of the candle. This is what happens when you look at an object through a magnifying glass. 5. Investigate: Return the candle to -24 and the left focal point to -12. Under Show specific line segments, turn off Apparent light lines. On the menu at lower right, select Concave lens. A concave lens curves inward on both sides. A. What do you notice about the three lines after they pass through the concave lens? On the opposite side of the lens from the object, the light rays diverge rather than converge at the focal point. Because a concave lens causes light rays to spread apart, it is also called a “diverging lens.” B. Turn on Apparent light lines. Is the image of the candle a real image or a virtual image? Explain. (Hint: Recall that a real image forms where light rays are focused.) Since the image is on the same side of the lens as the object, it is a virtual image because it does not originate where the light rays are truly focused. C. Move the candle back and forth. No matter where the candle is placed, what is true about the image? The picture remains virtual. D. What would an observer see if she looked at the candle through the concave lens? Through the concave lens, the observer would only be able to see the smaller picture. 6. Summarize: What are the differences between the images formed by convex and concave lenses? Depending on the positioning of the item and the focus point, the image in a convex lens may be smaller or larger than the actual object. The image is actual and inverted when the subject is behind the focal point. The picture is virtual, upright, and larger than the object when the subject is in line with the focal point and the lens. The picture will always be virtual, smaller than the object, and upright in a concave lens. Get the Gizmo ready: Activity B: ● Select the Convex lens. ● Move the candle to -15 and the focal point to -10. ● Turn off all line segments, and turn on Show ruler. Thin-lens equation Question: How is the position of the image related to the position of the object and the focal length of the lens? 1. Measure: In this activity, you will measure the relationships between several values: do: Distance between object and lens f: Distance between focal points and lens di: Distance between image and lens What are the current values of these variables? (Note: If you like, use the ruler to measure di.) do = 15 f= 10 di = 30 2. Gather data: Use the ruler to measure di for each of the following values of do and f. For the last three rows of the table, use your own values of do and f. do f di 15 10 30 1/15 1/30 1/10 25 10 16 1/16 1/16 1/10 20 10 1/20 1/20 1/10 29 10 20 15 1/29 1/15 1/10 30 15 30 1/30 1/30 1/15 3. Calculate: Find the reciprocal of each value and fill in the last three columns of the table. This is shown in the table above. 4. Analyze: For each row of the table, find the sum of and . 1/15+1/30=1/10 What do you notice? 1/25+1/16=1/10 1/20+1/20=1/10 1/30+1/30=1/15 1/do+1/di=1/f 5. Make a rule: Express the relationship between 1/do+1/di = , , and as an equation. 1/f This equation is called the thin lens equation because it assumes a very thin lens. Errors are introduced when a thicker lens is used. 6. Practice: A candle is placed 6 cm in front of a convex lens. The image of the candle is focused on a sheet of paper that is exactly 12 cm behind the lens. What is the focal length of the lens? 4 cm Show your work: 1/do+1/di=1/f ⅙+1/12=1/f ¼=1/f 4=f f=4 7. Practice: A candle is placed in front of a convex lens with a focal length of 5 cm. The image of the candle is focused on a sheet of paper that is exactly 15 cm behind the lens. What is the distance from the candle to the lens? 7.5 cm Show your work: 1/do + 1/di = 1/f 1/do + 1/15 = 1/5 1/do = 1/5 -1/15 1/do = 2/15 do = 7.5 8. On your own: Use the Gizmo to determine if the thin-lens equation applies to a concave lens. Report the results of your investigation below. (Hint: If the image is virtual, di and f are negative.) Concave lenses can be used with the thin lens equation since they still provide the right distance for the corresponding variable. However, you must make sure that di and f are both negative when added to the thin lens equation if the picture in these concave lenses is virtual. Activity C: Magnification Get the Gizmo ready: ● Select the Convex lens and turn on Show ruler. ● Move the candle to -15 and the focal point to -10. ● Adjust the height of the candle to 5.0 units. Question: What determines the magnification of an image? 1. Measure: What are the current values of these variables? do = 15 f= 10 di = 30 2. Measure: In addition to the variables you explored in the previous activity, you will now investigate several others. Measure each of these variables and list their values below. ho: Object height hi: Image height so: Object-focal point distance si: Image-focal point distance ho = 5 hi = 10 so = 5 si = 20 3. Gather data: Use the values above to fill in the first row of the table. Then run your own experiments to fill in the last three rows: do f di ho hi so si 15 10 30 5 10 5 20 2 2 2 2 20 10 20 5 5 10 10 1 1 1 1 25 11 20 5 4 14 9 0.8 0.8 0.8 0.8 30 10 15 10 5 20 5 0.5 0.5 0.5 0.5 4. Calculate: Calculate the given ratios to fill in the last four columns of the table. 5. Analyze: What do you notice about the four ratios? All of them are the same. 6. Make a rule: The magnification of an image is equal to the ratio of the image height to the object height. Using the ratios from the table on the previous page, write three equations that could be used to calculate magnification: 7. Practice: A candle is placed 4 cm in front of a convex lens. The image of the candle is focused on a sheet of paper that is exactly 10 cm behind the lens. What is the magnification of the image? -2.5 Show your work: M = -di / do M = -10/4 M = -2.5 8. Practice: A 29-cm pencil is placed 35 cm in front of a convex lens and illuminated by a spotlight. The focal point of the lens is 28 cm from the lens. A. What is the height of the pencil’s image? 116 cm B. How far from the lens will the image of the pencil be focused? 140 cm Show your work: a) hi / ho = f / so hi / 29 = 28/7 hi / 29 = 4 hi = 4 x 29 hi = 116 b) hi / ho = di / do 116 / 29 = di / 35 4 = di / 35 4 x 35 = di 140 = di di = 140 9. Challenge: When the object is between the focal point and the lens, so is negative. What other values must be negative for your equations to still work? Explain. (Hint: The distance between the focal point and lens (f) is never negative unless the lens is concave.) Another value that must be negative is di because, if the object is between the focus point and the lens, di must also be negative because, as the picture is refracted to the same side as the object, it becomes a VIRTUAL rather than a REAL image. Given that justification, it follows that the hi and si must also be negative.