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InClass

High School. PHYSICS. InClass. by SSL Technologies. with S. Lancione. Exercise-47. Plane mirrors. Plane Mirrors. THE LAW OF REFLECTION When light strikes a surface and is reflected, it changes direction. The direction it takes depends upon the angle it strikes the surface.

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InClass

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  1. High School PHYSICS InClass by SSL Technologies with S. Lancione Exercise-47 Plane mirrors

  2. Plane Mirrors THE LAW OF REFLECTION When light strikes a surface and is reflected, it changes direction. The direction it takes depends upon the angle it strikes the surface. As illustrated below, a ray of light going towards a surface is called an incident ray while a ray of light which is reflected away from a source is called a reflected ray. The law of reflection says that the angle of the incident ray with the normal equals the angle of the reflected ray (also with the normal). Click

  3. Plane Mirrors By definition, the angle which the incident ray makes with the normal is called the angle of incidence while the angle which the reflected ray makes with the normal is called the angle of reflection. Note that the incident ray, the reflected ray, and the normal are all on the same plane. Remember: The angle of incidence (i) is the angle formed by the incident ray and the normal(not the reflecting surface). The angle of reflection (r) is the angle between the reflected ray and the normal. Click

  4. Plane Mirrors The law of reflection, therefore, simply states that when a ray of light is reflected from a surface, it is reflected in such a direction that the incident angle equals the reflected angle. Using the law of reflection, we can determine the location of images formed by plane mirrors. Click

  5. Normal NOTE Since the two reflected rays diverge, we must “extend” them so that they will intersect. There are two steps in finding the image of a plane mirror. Normal Where the two reflected rays meet is the image point of the object point. Since we are not given the location of the object, we can select any arbitrary location. Step-1 Such asright here! Step-2 Draw an incident ray slightly upwards towards the mirror. Then draw its reflected ray in accordance with the law of reflection. Draw an incident ray slightly downwards towards the mirror, then draw its reflected ray. TASK Rather than draw an actual object, it is easier to draw the object as an arrow.This way, we can tell if theimage is inverted or not. An object is placed in frontof a plane mirror. Find the location and characteristics of the image. Reflected ray We can now draw the image. Incident ray Extended ray Extended ray Incident ray Objects consist of an infinite number of points. And each point has an infinite number of rays radiating outwards. Images also consist of an infinite number of points. Each point on the object has a corresponding point on the image. In locating an image, we take one point on the object and find its corresponding point (location) on the image object. As shown above, we usually take the (extreme) top point and find its corresponding point on the image object. Like this Image Reflected ray Object Back of mirror Click Click Click Click Click Click Click Click Click Click Click Click Click Click

  6. REMEMBER Virtual images are formed by extended rays. Real images are formed by reflected rays. Click

  7. Field of vision

  8. Mirror NOTE There are two steps in finding the field of vision. angle of incidence = angle of reflection angle of incidence = angle of reflection Step-2 Step-1 Draw a ray from the eye to the other end of the mirror. This is the reflected ray. Then, in accordance with the law of reflection, draw the incident ray. Draw a ray from the eye to one end of the mirror. This is the reflected ray. Then, in accordance with the law of reflection, draw the incident ray. Eye TASK Given a plane mirror and the position of an observer’s eye, determine the field of vision. Like this Like this Reflected ray Incident ray Normal Reflected ray Normal We can now draw the field of vision. It is the area bounded by the mirror and the two incidentrays. Incident ray Field of vision (Any point within this area will be reflectedto the observer’s eye as an image point.) Click

  9. TASK Given a plane mirror, a point object (P) and the position of an observer’s eye, determine whether or not the observer can see object point P. Mirror NOTE To determine whether or not the observer can see the given object, we first locate the image of the object. Next, we draw a line from the image point to the observer (eye). If the line intersects the mirror, the observer can see the image. If the line does not intersect the mirror, then the observer cannot see the image. Perpendicular to mirror Look! The line does not intersect the mirror. Conclusion: observer cannot see the image. Object point Image point Eye P' Like this Image distance Next, since the image distance equals the object distance, we mark off the image point, P‘, along the extended line. To find the location of the image point, we start by drawing a line perpendicular to the mirror and extend it beyond the mirror. Object distance = Image distance Object distance ANSWER The observer cannot see the image. Finally, we draw a straight line from the observer’s eye to the image point. If this line intersects the mirror, the observer can see the image. If this line does not intersect the mirror, the observer cannot see the image. P Click

  10. Image point Mirror NOTE To determine whether or not the observer can see the given object, we first locate the image of the object. Next, we draw a line from the image point to the observer’s eye. If the line intersects the mirror, the observer can see the image. If the line does not intersect the mirror, then the observer cannot see the image. Perpendicular to mirror Look! The line does not intersect the mirror. Thus, the observer cannot see the image. Object point Eye P' TASK Given a plane mirror, a point object (P) and the position of an observer’s eye, determine whether or not the observer can see object P. Image distance ANSWER The observer cannot see the image. We now draw a straight line from the observer’s eye to the image point. If this line intersects the mirror, the observer can see the image. If this line does not intersect the mirror, the observer cannot see the image. To find the location of the image point, we start by drawing a line perpendicular to the mirror and extend it beyond the mirror. Object distance = Image distance Object distance P Click

  11. EXERCISES

  12. Question-1 State the Law of Reflection. The angle of incidence equals the angle of reflection. Click Click Click

  13. Question-2 The diagram below represents an object in front of a plane mirror. Object Note: Diagram not drawn to scale. a) Draw the image. 4 cm b) How high is the image? 20 cm c) How far is the image from the object? Click Click Click Click Click

  14. Question-3 Explain why there is only a lateral(left-right) reversal when we look at ourselves in a plane mirror. Because when we turn towards a mirror, we do so by turning aboutthe Y-axis (left-right) and not about the X-axis (top-bottom). Click Click

  15. Question-4 The diagram below illustrates the image of an object produced by a plane mirror. Label the incident ray, the reflected ray and the extended ray. Reflected ray Extended ray Incident ray Click Click Click

  16. Question-5 Explain how realimages are formed and how virtualimages are formed. Realimages are formed by the intersection of reflectedrays, virtualimages are formed by the intersection of extendedrays. NOTE As long as there is at least one extended ray,the image is said to be virtual. Real images are formed only by reflected rays. Click Click

  17. Question-6 Half Stefania’s height Stefania is 1.5 m tall. Prove that the shortest mirror necessaryfor Stefania to see her full height is 75 cm (half her height). Divide Stefania’s height into two parts, from her eyes to the top ofher head and from her eyes to the bottom of her feet. Thus, in order for a person to see their full height in a plane mirror,the minimum mirror required is a mirror half their height. Click Click

  18. Question-7 A beam of light is reflected from a plane mirror such that the angle between the incident ray and reflected ray is 50o. Draw the beam and calculate the angle of incidence? . Click Click

  19. Question-8 Two mirrors, M1 and M2, are at 90o to each other. As illustrated, a beam of light strikes mirror M1 with an angle of incidence of 60o and is reflected by mirror M2. Complete the diagram and find theangle of incidence of the ray reflected by mirror M2. Angle of incidence of mirror M2 Click Click

  20. Question-9 Two plane mirrors, M1 and M2, are at 60o to each other as illustrated in the diagram below. A beam of light strikes mirror M1 with an angle of incidence of 40o. Complete the diagram and determine the angle of reflection of the beam reflected from mirror M2 ? Angle of reflection of mirror M2 Click Click

  21. Question-10 Two mirrors are parallel to each other as illustrated in the diagram below. A beam of light strikes the beginning of one mirror at an angle of incidence of 35o. Complete the diagram and find the number of times the beam is reflectedbefore it emerges fromthe two mirrors? . Click Click

  22. Question-11 Angle of reflection Angle of incidence What is the angle of reflection for the ray diagram illustrated below? REMINDERAll angles are measured from the “normal”. A) 1 B) 2 C) 3 D) 4 Click

  23. Question-12 Answer: Stefania should choose theconvex mirror in order toobtain the maximum field of view. Stefania wants to install a rear view mirror on her bike. She has achoice of choosing a plane or a convex mirror as illustrated below. Which mirror should Stefania choose in order to obtain the maximum filed of view? Observer Observer Click

  24. Question-13 The following set up consists of a light source, an obstacle,six different locations (labeled 1 to 6) and a target. At which one of the locations can you place a plane mirrorso that the light source is reflected to the target? Note that only at location 5 canwe draw unobstructed lines bothto the source and to the target. Thus, we must place the mirrorat location 5. To determine the correct location,we draw two lines from each point. One line to the light source andanother line to the target. If bothlines are unobstructed, that is thepoint where we place the mirror. We now divide the angle between the two lines at location 5 with a bisector. This bisector will be the normal to the mirror. Finally, we draw the mirror perpendicular to the normal (bisector). If required, using a protractor, we can measure the angle of incidence or the angle of reflection. No No No No No Yes Mirror Click

  25. Question-14 The set up below consists of an object and its shadow cast ona screen by a point light source. This problem illustrates the fact that light travels in a straight line, technically knownas rectilinear propagation. Determine the position of the light source. A) Point-A Draw a line from the top of the shadowto the top of the object and then extendthe line to the light source. B) Point-B C) Point-C D) Point-D Click Click

  26. SSLTechnologies.com/science The end

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