HW 9 Ch 23 Ray OpticsDue: 11:59pm on Tuesday, November 17, 2015 To understand how points are awarded, read the Grading Policy for this assignment. Ray Tracing for a Concave Mirror An object O is placed at the location shown in front of a concave spherical mirror. Use ray tracing to determine the location and size of the reflected image. As you work, keep in mind the following properties of principal rays: 1. A ray parallel to the axis, after reflection, passes through the focal point F of a concave mirror or appears to come from the (virtual) focal point of a convex mirror. 2. A ray through (or proceeding toward) the focal point F is reflected parallel to the axis. 3. A ray along the radius through or away from the center of curvature C intersects the surface normally and is reflected back along its original path. 4. A ray to the vertex V is reflected, forming equal angles with the optic axis. Part A Trace the path of a ray emitted from the tip of the object through the focal point of the mirror and then the reflected ray that results. Start by extending the existing ray emitted from the tip of the object. Then create the reflected ray. Draw the vector for the reflected ray starting from the point where the incident focal ray hits the mirror. The location and orientation of the vector will be graded. The length of the vector will not be graded. ANSWER: Correct Part B Now trace the path of a ray emitted from the tip of the object parallel to the axis of the mirror and then the reflected ray that results. Start by extending the existing ray emitted from the tip of the object. Then create the reflected ray. Draw the vector for the reflected ray starting from the point where the incident focal ray hits the mirror. The location and orientation of the vector will be graded. The length of the vector will not be graded. ANSWER: Correct Part C Now, using the two rays you have traced, draw the image of the object at the correct location, with the correct orientation and size. Draw the vector ending at the intersection of the reflected rays. The location, orientation, and length of the vector will be graded. Hint 1. Image formation (tip of the arrow) Since both the focal ray and parallel ray reflect from the tip of the object, where they cross is the location of the image of the tip of the object. Hint 2. Image formation (base of the arrow) If you were to draw the focal ray and parallel ray from the base of the object, which lies on the central axis of the mirror, these rays would both travel toward and reflect from the mirror along the central axis. This implies that if the base of the object lies on the central axis, the base of the image lies on the central axis. ANSWER: Correct You could have drawn additional rays such as the ray incident from the tip of the object that strikes the vertex of the mirror and its corresponding reflected ray. However, two rays are often enough to determine the image location. Ray Tracing for a Convex Mirror An object O is placed at the location shown in front of a convex spherical mirror. Use ray tracing to determine the location and size of the image in the mirror. As you work, keep in mind the following properties of principal rays: 1. A ray parallel to the axis, after reflection, passes through the focal point F of a concave mirror or appears to come from the (virtual) focal point of a convex mirror. 2. A ray through (or proceeding toward) the focal point F is reflected parallel to the axis. 3. A ray along the radius through or away from the center of curvature C intersects the surface normally and is reflected back along its original path. 4. A ray to the vertex V is reflected, forming equal angles with the optic axis. Part A Trace the path of a ray emitted from the tip of the object toward the focal point of the mirror and then the reflected ray that results. Start by extending the existing ray emitted from the tip of the object. Then create the reflected ray. Draw the vector for the reflected ray starting from the point where the incident focal ray hits the mirror. The location and orientation of the vector will be graded. The length of the vector will not be graded. ANSWER: Correct Part B Now trace the path of a ray emitted from the tip of the object parallel to the axis of the mirror and then the reflected ray that results. Start by extending the existing ray emitted from the tip of the object. Then create the reflected ray. Draw the vector for the reflected ray starting from the point where the incident focal ray hits the mirror. The location and orientation of the vector will be graded. The length of the vector will not be graded. Hint 1. Constructing the reflected ray According to point #1 in the problem introduction, the reflected ray will appear to have originated from the focal point. Try drawing a vector from the focal point through the point where the incident ray hits the mirror. Then shorten this vector so that its start point is on the mirror, without changing its angle. Alternatively, you can draw an unlabeled vector extending from the focal point and passing through the point where the incident ray hits the mirror. Then you can draw the reflected ray on top of the unlabeled vector. Keep in mind that the unlabeled vector should not be part of your submission. ANSWER: Correct Part C What type of image of the object will the convex mirror create? ANSWER: A real image in front of the mirror. A virtual image behind the mirror. Correct Part D Now, using the two rays you have traced, draw the image of the object at the correct location, with the correct orientation and size. Start by extending the existing virtual rays from the surface of the mirror where the reflected rays touch the mirror. Draw the vector ending at the intersection of the virtual rays. The location, orientation, and length of the vector will be graded. Hint 1. Virtual extension A virtual extension is a ray on the virtual side of the mirror that represents the path that the reflected ray appears to have followed inside the mirror. To draw a virtual extension, simply extend the virtual ray on the virtual side of the mirror without changing its direction. Hint 2. Image formation (tip of the arrow) Since both the focal ray and parallel ray reflect from the tip of the object, where they cross is the location of the image of the tip of the object. Hint 3. Image formation (base of the arrow) If you were to draw the focal ray and parallel ray from the base of the object, which lies on the central axis of the mirror, these rays would both travel toward and reflect from the mirror along the central axis. This implies that if the base of the object lies on the central axis, the base of the image lies on the central axis. ANSWER: Correct The simple ray tracing you worked through in this problem is the starting point for the photorealistic scenery and lighting effects that are common in computer games. Reflection Vector Drawing Vittorio needs to clean the dusty mirror hanging on the wall before he can use it, but he would like to clean as little of the mirror as possible. Part A Vittorio would like to be able to see the logo on his shirt. Draw the incident and reflected rays showing the light from the logo reflecting off the mirror into his eyes. The rays should meet at the point on the mirror that needs cleaning. Adjust the existing vectors on the diagram, for the reflected (R) and incident (I ) ray, starting and ending at the surface of the mirror respectively. The location and orientation of the vectors will be graded. Hint 1. The law of reflection The angle an incident light ray makes with a mirror is the same as the angle made by the reflected ray. Hint 2. Equal angles and equal distances Since the objects Vittorio is interested in seeing are at (approximately) the same distance from the mirror as his eyes, the fact that the incident and reflected angles are the same means that the vertical distances traveled by the light on incidence and reflection are the same. ANSWER: Correct Part B Vittorio would like to be able to check whether his shoelaces are tied. Draw the incident and reflected rays showing the light from his shoelaces reflecting off the mirror into his eyes. Again, the rays should meet at the point on the mirror that needs cleaning. (See the hints from Part A if you need help.) Adjust the existing vectors on the diagram, for the reflected (R) and incident (I ) ray, starting and ending at the surface of the mirror respectively. The location and orientation of the vectors will be graded. ANSWER: Correct Images Produced by Different Concave Mirrors Conceptual Question d f The same object is placed at different distances d in front of six different concave spherical mirrors. Each mirror has the focal length f listed below. Part A Which, if any, of these scenarios produce a real image? Which, if any, of these scenarios produce a virtual image? Sort the following scenarios into the appropriate bins. Hint 1. How to draw a principal ray diagram The properties of the image can be determined graphically by drawing a principal ray diagram. Start by drawing two rays from the object, coming from the same point. Draw one ray parallel to the optical axis; its reflected ray will go through the focal point. Draw the other ray going through the focal point; its reflected ray will be parallel to the optical axis. The image will be formed at the point where the two rays intersect. (You might need to extend the reflected rays straight back through the mirror to find the intersection point.) Hint 2. Definition of a real image A real image is an image that is formed when light rays originating from an object are reflected off the mirror and intersect. If the light rays from the object intersect only when they are extended through the mirror, then the image formed is virtual. Hint 3. Principal rays for objects within the focal length Hint 4. Principal rays for objects outside the focal length ANSWER: Correct Part B Which, if any, of these scenarios produce an inverted image? Which, if any, of these scenarios produce an upright image? Sort the following scenarios into the appropriate bins. ANSWER: Correct Part C Rank the images on the basis of the magnitude of their magnification. Rank these from largest to smallest. To rank items as equivalent, overlap them. Hint 1. Ratio of focal length to object distance from focal point You can determine the magnification of the image using the object distance, d , and focal length, f . The ratio of the focal length to object distance from the focal point is all you need. Recall that the object distance, d , is measured from the plane of the mirror. A sketch of the mirror will help you to visualize the distances given and ratio you require. Hint 2. Object at the focal point If an object is located exactly at the focal point of the mirror, d point, in either direction, the image size decreases. = f , the image is "infinitely" large. As the object moves away from the focal ANSWER: Correct Reflection and Refraction Ranking Task A ray of light is incident onto the interface between material 1 and material 2. Part A Given the indices of refraction n 1 and n 2 of material 1 and material 2, respectively, rank these scenarios on the basis of the phase shift in the refracted ray. Rank from largest to smallest. To rank items as equivalent, overlap them. Hint 1. Distinguish between reflection and refraction The light that reflects off of the interface does so in a similar manner to light reflecting from a mirror. The light that refracts through the interface has its path bent by the differing indices of refraction of the two materials. A phase shift occurs in only one of these two processes. Which one? ANSWER: reflection refraction Correct ANSWER: Correct Part B Rank these scenarios on the basis of the phase shift in the reflected ray. Rank from largest to smallest. To rank items as equivalent, overlap them. Hint 1. Phase change upon reflection Reflected light will experience a 180degree phase change when it reflects from a medium of higher index of refraction and no phase change when it reflects from a medium of smaller index. The amount of the phase change is independent of the magnitude of the difference between the two indices of refraction. ANSWER: Correct Ray Tracing and Image Formation with a Concave Lens A concave lens refracts parallel rays in such a way that they are bent away from the axis of the lens. For this reason, a concave lens is referred to as a diverging lens. Part A Consider the following diagrams, where F represents the focal point of a concave lens. In these diagrams, the image formed by the lens is obtained using the ray tracing technique. Which diagrams are accurate? Type A if you think that only diagram A is correct, type AB if you think that only diagrams A and B are correct, and so on. Hint 1. A ray parallel to the lens axis A ray parallel to the axis of a concave lens is refracted along a line that extends back through the focal point on the same side of the lens. Hint 2. A ray that passes through the focal point A ray that is directed toward the focal point on the other side of the lens is refracted parallel to the lens axis. Hint 3. A ray that passes through the middle of the lens A ray that passes through the middle of a concave lens continues on its original direction with essentially no displacement after passing through the lens. ANSWER: AC Correct A concave lens always forms an image that is on the same side of the lens as the object. Part B If the focal length of the concave lens is 7.50 cm , at what distance d o from the lens should an object be placed so that its image is formed 3.70 cm from the lens? Express your answer in centimeters. Hint 1. How to approach the problem To determine the object distance you can use the thinlens equation, but be careful to assign the correct sign to each quantity involved in the equation. Hint 2. The thinlens equation The thinlens equation for a lens with a focal length f is 1 do + 1 di = 1 f , where d o and d i are the object distance and the image distance, respectively. Hint 3. Find the image distance What is the image distance d i for a concave lens that forms an image 3.70 cm from the lens? Express your answer in centimeters. Hint 1. Sign convention for image distances Conventionally the image distance has a positive sign when the image is on the opposite side of the lens from the object (a real image), and a negative sign when the image is on the same side of the lens as the object (a virtual image). ANSWER: di = cm ANSWER: do = 7.30 cm Correct Part C What is the magnification m produced by the concave lens described in Part B? Express your answer numerically. Hint 1. Magnification The magnification m produced by a lens is given by the formula m=− di do , where d o and d i are the object distance and the image distance, respectively. ANSWER: m = 0.507 Correct Part D Where should the object be moved to have a larger magnification? Hint 1. Magnification and image size. Recall that a larger magnification corresponds to a larger image. You may find the ray diagrams found in Part A helpful in determining how the size of the image varies as the object is moved closer or farther from the lens. ANSWER: The object should be moved closer to the lens. The object should be moved farther from the lens. The object should be moved to the focal point of the lens. The object should not be moved closer to the lens than the focal point. Correct Tactics Box 23.2 Ray Tracing for a Converging Lens Learning Goal: To practice Tactics Box 23.2 Ray Tracing for a Converging Lens. The procedure known as ray tracing is a pictorial method for understanding image formation when lenses or mirrors are used. It consists of locating the image by the use of just three "special rays." The following tactics box explains this procedure for the case of a converging lens. TACTICS BOX 23.2 Ray tracing for a converging lens A. Draw an optical axis. Use graph paper or a ruler. Establish an appropriate scale. B. Center the lens on the axis. Mark and label the focal points at distance f on either side. C. Represent the object with an upright arrow at distance s . It is usually best to place the base of the arrow on the axis and to draw the arrow about half the radius of the lens. D. Draw the three "special rays" from the tip of the arrow. Use a straightedge. 1. A ray parallel to the axis (Ray 1) refracts through the far focal point. 2. A ray that enters the lens along a line through the near focal point (Ray 2) emerges parallel to the axis. 3. A ray through the center of the lens (Ray 3) does not bend. E. Extend the rays until they converge. This is the image point. Draw the rest of the image in the image plane. If the base of the object is on the axis, then the base of the image will also be on the axis. F. Measure the image distance s ′ . Also, if needed, measure the image height relative to the object height. Follow the steps above to solve the following problem: An object is 9.0 cm from a converging lens with a focal length of 2.8 cm. Use ray tracing to determine the location of the image. Part A The diagram below shows the situation described in the problem. The focal length of the lens is labeled f ; the scale on the optical axis is in centimeters. Draw the three special rays, Ray 1, Ray 2, and Ray 3, as described in the tactics box above, and label each ray accordingly. Do not draw the refracted rays. Draw the rays from the tip of the object to the lens plane. ANSWER: Correct Part B Now, draw the refracted segments of the three special rays considered previously. Use the labels Ray1r , Ray2r , and Ray3r for the refracted segment of Ray 1, the refracted segment of Ray 2, and the refracted segment of Ray 3, respectively. Make sure to extend the refracted rays until they all converge. If your rays do not all converge at the same point, you may need to be more precise in your drawing. ANSWER: Correct Part C Based on the ray diagram drawn above, at what distance s ′ from the lens (plane) does the image form? ANSWER: at approximately 2 cm at approximately 4 cm at approximately 6 cm Correct Tactics Box 23.3 Ray Tracing for a Diverging Lens Learning Goal: To practice Tactics Box 23.3 Ray Tracing for a Diverging Lens. The procedure known as ray tracing is a pictorial method of understanding image formation when lenses or mirrors are used. It consists of locating the image by the use of just three "special rays." The following tactics box explains this procedure for the case of a diverging lens. TACTICS BOX 23.3 Ray tracing for a diverging lens 1. Draw an optical axis. Use graph paper or a ruler. Establish an appropriate scale. 2. Center the lens on the axis. Mark and label the focal points at distance f on either side. 3. Represent the object with an upright arrow at distance s . It is usually best to place the base of the arrow on the axis and to draw the arrow about half the radius of the lens. 4. Draw the three "special rays" from the tip of the arrow. Use a straightedge. 1. A ray parallel to the axis (Ray 1) diverges along a line through the near focal point. 2. A ray along a line toward the far focal point (Ray 2) emerges parallel to the axis. 3. A ray through the center of the lens (Ray 3) does not bend. 5. Trace the diverging rays' tails backward. The point from which they are diverging is the image point, which is always a virtual image. 6. Measure the image distance s ′ . Also, if needed, measure the image height relative to the object height. Follow the steps above to solve the following problem: An object is 9.0 cm from a diverging lens with a focal length of 5.2 cm. Use ray tracing to determine the location of the image. Part A The diagram below shows the situation described in the problem. The focal length of the lens is labeled f ; the scale on the optical axis is in centimeters. Draw the three special rays, Ray 1, Ray 2, and Ray 3, as described in the tactics box above, and label each ray accordingly. Do not draw the refracted rays. Remember to use the crosshairs to position the start point and end point of your rays. Draw the rays from the tip of the object to the central axis of the lens. ANSWER: Correct Part B Now, draw the refracted rays corresponding to the three special rays considered previously. Use the labels Ray1r , Ray2r , and Ray3r for the refracted segment of Ray 1, the refracted segment of Ray 2, and the refracted segment of Ray 3, respectively. Make sure to trace the refracted rays' tails backward to the point from which they diverge. ANSWER: Correct Part C Based on the rays drawn above, at what distance s ′ from the lens will the image form? ANSWER: at approximately 3.5 cm on the opposite side of the lens at approximately 3.5 cm on the same side of the lens at approximately 5 cm on the opposite side of the lens at approximately 5 cm on the same side of the lens Correct Tactics Box 23.4 Ray Tracing for a Spherical Mirror Learning Goal: To practice Tactics Box 23.4 Ray tracing for a spherical mirror. The procedure known as ray tracing is a pictorial method of understanding image formation when lenses or mirrors are used. It consists in locating the image by the use of just three "special rays." The following Tactics Box explains this procedure for the case of a concave mirror. TACTICS BOX 23.4 Ray tracing for a spherical mirror A. Draw an optical axis. Use graph paper and a ruler. Establish an appropriate scale. B. Center the mirror on the axis. Mark and label the focal point at distance f from the mirror's surface. C. Represent the object with an upright arrow at distance s . It is usually best to place the base of the arrow on the axis and to draw the arrow about half the radius of the mirror. D. Draw the three "special rays" from the tip of the arrow. Use a straightedge. 1. A ray parallel to the axis (Ray 1) reflects through (concave) or away from (convex) the focal point. 2. An incoming ray passing through (concave) or away from (convex) the focal point (Ray 2) emerges parallel to the axis. 3. A ray that strikes the center of the mirror (Ray 3) reflects at an equal angle on the opposite side of the optical axis. E. Extend the rays until they converge. This is the image point. Draw the rest of the image in the image plane. If the base of the object is on the axis, then the base of the image will also be on the axis. F. Measure the image distance s ′ . Also, if needed, measure the image height relative to the object height. Follow the steps above to solve the problem: An object is 18.0 cm from a concave mirror with a focal length of 7.5 cm. Use ray tracing to determine the location of the image. Part A The diagram below shows the situation described in the problem. The focal length of the mirror is labeled f ; the scale on the optical axis is in centimeters. Draw the three special rays Ray1, Ray2, and Ray3 as described in the Tactics Box above, and label each ray accordingly. Draw the rays from the tip of the object to the mirror. Do not draw the reflected rays. ANSWER: Correct Part B Now draw the reflected segments of the three special rays considered previously. Use the label Ray1r for the reflected segment of Ray 1, Ray2r for the reflected segment of Ray 2, and so forth. Make certain to extend the reflected rays until they converge. ANSWER: Correct Part C Based on the rays drawn above, at what distance s ′ from the mirror does the image form? ANSWER: at about 5.5 cm at about 7.5 cm at about 13 cm Correct PhET Tutorial: Geometric Optics Learning Goal: To understand how the properties of a convex lens and the distance between the object and the lens affect the image. For this tutorial, use the PhET simulation Geometric Optics. This simulation allows you to experiment with properties of a convex lens and see how the resulting image changes. Start the simulation. When you click the simulation link, you may be asked whether to run, open, or save the file. Choose to run or open it. You can drag the lens and the object to any location and see the resulting image. You can adjust the curvature of the lens, the refractive index of the lens material, and the diameter of the lens, using the slider bars at the top. Feel free to play around with the simulation. When you are done, begin part A. Part A A convex lens is a transparent instrument that uses refraction to bend and focus light from an object, forming a sharp image. We'll first investigate how the lens produces an image from a point source of light (a lamp). Select Screen at upper right, which makes the object a lamp and gives a black screen that can be dragged around. Select Many rays at upper left to see how the lens bends some of the rays of light from the lamp. With the lamp positioned far to the left of the lens, you should see that the rays that go through the lens converge to a point. If the screen is placed where the beams converge, the image on the screen will be in focus (it will be a small dot of light since that is what the object looks like in this case). The screen is then at the focal plane. As the lamp is moved closer to the lens, the distance between the focal plane and the lens _____. ANSWER: remains the same. decreases. increases. Correct When the lamp is closer to the lens, the rays going into the lens are diverging more quickly than when the lamp is further away. Thus, after going through the lens, they aren't converging as quickly, so it takes a longer distance for them to focus. Part B The horizontal blue line through the middle of the lens is called the optical axis. As the lamp is moved above or below the axis, keeping the horizontal distance to the center of the lens constant, how does the horizontal distance from the lens to the image change? ANSWER: It does not change. It increases. It decreases. Correct This explains why we refer to a focal plane. All objects with the same horizontal distance to the lens will be focused by the lens on the same plane. You can see this by selecting 2nd Point and placing one of the lamps above original lamp. Part C Do the rays that go through the lens always converge on the right side of the lens (forming a real image), regardless of the position of the lamp? ANSWER: No; if the lamp is very far away, the rays don't converge. Yes No; if the lamp is very close to the lens, the rays don't converge. Correct The two yellow X's (on either side of the lens) are called focal points. The distance between a focal point and the lens is called the focal length of the lens. If the object is closer to the lens than this distance, the lens cannot bend the rays enough for them to converge. Part D Place the lamp so that the pointlike source is directly on the left focal point. What happens to the rays that pass through the lens on the right side of the lens? ANSWER: They are parallel. They diverge. They converge. Correct When the object is at the focal point, the rays neither diverge nor converge, but are perfectly parallel. Conversely, if the object were at infinity, the rays going from the object to the lens would be parallel, and after passing through the lens, they would converge at the focal point. Part E The focal length ____________ when the refractive index of the lens is increased and __________ when the curvature radius of the lens is increased. ANSWER: decreases / increases increases / decreases decreases / decreases increases / increases Correct A higher index of refraction causes the rays of light to bend more (going into and coming out of the lens). The rays therefore focus more quickly, and the focal length is shorter. If the surface of the lens where the ray enters is parallel to the surface where the ray exits, the ray will not be bent at all. The greater the angle between the two surfaces where the ray enters and leaves the lens (as is the case for a smaller radius of curvature), the more the outgoing ray will be bent relative to the incoming ray. Part F The focal length __________ when the diameter of the lens is increased. ANSWER: increases does not change decreases Correct The focal length does not depend on the lens's diameter; a greater diameter simply allows more light to be focused. Part G Make the curvature radius 0.6 m, the refractive index 1.5, and the diameter 0.6 m. Place the lamp so that the source of light is 120 cm from the middle of the lens (use the ruler). The focal length of the lens is ____________, and the focal plane is ___________ from the lens. Hint 1. How to approach the problem Use the ruler to measure the distance from the center of the lens to either of the focal points. This is the focal length. Then, use the ruler to measure the distance from the center of the lens to where the rays converge (the location of the image). ANSWER: 30 cm / 120 cm 120 cm / 120 cm 60 cm / 60 cm 60 cm / 180 cm 60 cm / 120 cm Correct For this lens, when the object is 120 cm away, the image is the same distance on the other side of the lens. Part H Now, move the lamp so that it is 90 cm from the center of the lens. How far from the lens is the image? ANSWER: 120 cm 160 cm 180 cm 60 cm 90 cm Correct The distances to the object, s ; the image, s ′ ; and the focal length, f , are related by the lens equation: 1 s + 1 s ′ = 1 f . If s decreases, then s ′ must increase (as you found in Part A). Notice that, as you found in Part F, s ′ = s if s is twice the focal length of the lens (1/120 + 1/120 = 1/60). Part I Now, let's look at the images of extended objects. Deselect Screen. You should see a pencil as the object and its image. Every point on the pencil emits rays like a point source. Selecting Many rays shows rays from the pencil’s tip. Move the pencil around, and look at the resulting image. How does the size of the image depend on the position of the pencil (keep the distance greater than the focal length of the lens)? ANSWER: The size of the image decreases as the distance from the pencil to the lens decreases. The size of the image doesn't change. The size of the image increases as the distance from the pencil to the lens decreases. Correct The image gets very big as the object approaches the focal point. Part J The magnification of an object is defined as M = h ′ h , where h′ is the height of the image and h is the height of the object. If the image is inverted, then h is negative. ′ Place the pencil 90 cm from the lens. What is the magnification of the image (be sure the curvature radius is still 0.6 m and the refractive index is 1.5)? ANSWER: − 1.5 − 1.0 − 2.0 − 0.5 Correct Notice that the image is also twice as far from the lens as the object. Part K How far from the lens does the pencil need to be for the magnification M to be 1? ANSWER: 180 cm 60 cm 120 cm It's impossible! Correct Notice that, as you saw earlier for this lens, when the object is 120 cm away, the image is also 120 cm, so the ratio of the distances s and s ′ is s equal to one. In fact, as you might have guessed by now, the magnification can also be expressed as M = − s . ′ PhET Interactive Simulations University of Colorado http://phet.colorado.edu Problem 23.52 One of the contests at the school carnival is to throw a spear at an underwater target lying flat on the bottom of a pool. The water is 1.20 m deep. You're standing on a small stool that places your eyes 3.20 m above the bottom of the pool. As you look at the target, your gaze is 30∘ below horizontal. At what angle below horizontal should you throw the spear in order to hit the target? Part A Your raised arm brings the spear point to the level of your eyes as you throw it, and over this short distance you can assume that the spear travels in a straight line rather than a parabolic trajectory. Express your answer with the appropriate units. ANSWER: 35.5 ∘ Correct Problem 23.39 A red ball is placed at point A in the figure. Part A How many images are seen by an observer at point O? Express your answer as an integer. ANSWER: N = 3 images Correct Part B What are the (x; y) coordinates of each image? Express your answer using two significant figures. Give your answer in the form (x;y). If there is more than one answer, separate them by commas. Give your answer in meters. ANSWER: (1.0;2.0),(1.0;2.0),(1.0;2.0) Correct Problem 23.74 A sports photographer has a 130mmfocallength lens on his camera. The photographer wants to photograph a sprinter running straight away from him at 5.0 m/s . Part A What is the speed (in mm/s) of the sprinter's image at the instant the sprinter is 15 m in front of the lens? Express your answer using two significant figures. ANSWER: |v| = 74 mm/s All attempts used; correct answer withheld by instructor Score Summary: Your score on this assignment is 87.5%. You received 11.37 out of a possible total of 13 points.
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