1. What is the law of reflection?

Transcription

1 Name: Skill Sheet 7.A The Law of Reflection The law of reflection works perfectly with light and the smooth surface of a mirror. However, you can apply this law to other situations. For example, how would the law of reflection help you win a game of pool or pass a basketball to a friend on the court? In this skill sheet you will review the law of reflection and perform practice problems that utilize this law. Use a protractor to make your angles correct in your diagrams.. What is the law of reflection? The law of reflection states that when an object hits a surface, its angle of incidence will equal the angel of reflection. This is true when the object is light and the surface is a flat, smooth mirror. When the object and the surface are larger and lack smooth surfaces, the angles of incidence and reflection are close but not always exact. Nevertheless, this law is very helpful in performing activities like bouncing a ball to someone or in playing pool.. When we talk about angles of incidence and reflection, we often talk about the normal. The normal to a surface is an imaginary line that is perpendicular to the surface. a. Draw a diagram that shows a surface, with a normal line, and a ray of light hitting the surface at an angle of incidence of 60 degrees. b. In the diagram above, label the angle of reflection. How many degrees is this angle of reflection? 2. Light strikes a mirror s surface at 30 degrees to the normal. What will the angle of reflection be? 3. The angle made by the angle of incidence and angle of reflection for a ray of light hitting a mirror is 90 degrees. What are the measurements of each of these angles?

2 Skill Sheet 7.A The Law of Reflection 4. In a game of basketball, the ball is bounced (with no spin) toward a player at an angle of 40 degrees to the normal. What will the angle of reflection be? Draw a diagram that shows this play. Label the angles of incidence and reflection and the normal. 2. Playing pool Use a protractor to figure out the angles of incidence and reflection for the following problems.. Because a lot of her opponent s balls are in the way for a straight shot, Amy is planning to hit the cue ball off the side of the pool table so that it will hit the 8-ball into the corner pocket. In the diagram, show the angles of incidence and reflection for the path of the cue ball. How many degrees does each angle measure? 2. You and a friend are playing pool. You are playing solids and he is playing stripes. You have one ball left before you can try for the eight ball. Stripe balls are in the way. You plan on hitting the cue ball behind one of the stripe balls so that it will hit the solid ball and force it to follow the pathway shown in the diagram. Use your protractor to figure out what angles of incidence and reflection are needed at points A and B to get the solid ball into the far side pocket. 2

3 Name: Skill Sheet 7.B Refraction When light rays cross from one material into the other they bend. This bending of light rays is called refraction. This phenomenon is very important and useful. All kinds of optics, from your glasses to your camera lens to your binoculars use this principle. In this skill sheet we will understand this phenomenon and learn how to calculate the actual amount of bending as light goes from one material into the other. All we need to know is the properties of the materials and some simple geometry.. Introduction to refraction The principle is illustrated on the diagram. In this case we have air and water as the two materials where light travels. A light ray making an angle θ with the vertical hits the water surface. This ray will enter the water at a different angle from vertical θ 2. Once we know the path that a light ray takes as it enters the water we also have the solution of the problem where the light ray starts from the water and enters the air. A fish in the water does not appear to be where it actually is. A light ray that leaves the fish enters our eyes after it as been refracted as shown on the diagram. So, if you are a hunter trying to spear this fish you better know about this phenomenon or the fish will get away. Very early humans realized this phenomenon and adjusted their aim with great success. Here are two questions to consider:. Why does the light ray bend as it crosses from one material into another? 2. How much does it bend? The answer to the first question is related to the properties of the materials. For the purpose of refraction, the property of the material is represented by a number called the index of refraction which is represented by the symbol n. We call n i the index of refraction of the material from which the ray is coming (incident material) and n r the index of refraction of the material to which the ray is entering (refractive material). Also the corresponding angles are called θ i and θ r. The general rules of thumb are: If n i is less than n r, then θ r is less than θ i. If n i is greater than n r, then θ r is greater than θ i. The answer to the second question is provided by Snell s law which describes the relation between n i, n r, θ r, and θ i. In working with this formula, the incident and refractive angles are measured from the normal to the surface. Also assume that the surface between the two materials is smooth.

4 Skill Sheet 7.B Refraction 2. Example problems Solve the following problems using Snell s law. The first problem is done for you as an example.. Air has an index of refraction equal to.0 and glass has an index of refraction equal to.5. Light travels from air into glass with an angle of incidence θ i = 25. What is the refractive angle θ r? We can solve for θ r from the relation: n i sinθ i = n r sinθ r n r.0 sinθ r = --- sinθ i = sin25 =.5 n i ( 0.423) = and the angle θ r is given by the inverse sine of 0.285: θ r = sin (0.282) = 6.38 = 6 2. Light travels from glass into air. The incident angle θ i = 30. What is the angle of refraction? 3. Light travels from air into glass. The angle of refraction is θ i = 30. What is the angle of incidence? 4. Light travels from air into diamond. The index of refraction for diamond is 2.4 and the incident angle θ i = 30. What is the angle of refraction? 5. If the incident and refraction angles are 30 degrees and 45 degrees, respectively, which material has the larger index of refraction and what is the ratio of the refraction indices? 6. Air has an index of refraction equal to.0 and glass has an index of refraction equal to.5. Light travels from glass into air. Calculate the incident angle for which the refractive angle equals 90 degrees. 7. When the angle of refraction becomes 90 degrees we have a very special and interesting situation. The incident angle that corresponds to this case is called the critical angle. When the incident angle becomes greater that the critical angle the incident light will now be reflected at the surface. This is called total internal reflection. Calculate the critical angle (the angle of incidence) as we did on problem 6 for light travelling from diamond to air. 8. Calculate the critical angle of refraction for the water-air interface. The index of refraction for water is.33. 2

5 Name: Skill Sheet 7.2 Ray Diagrams Here you will get practice in making ray diagrams. A ray diagram helps you determine where an image produced by a lens will form and whether the image will be upside down or right side up. For each question on this skill sheet, read the directions carefully and plot your ray diagram in the space provided.. Getting started. Of the diagrams below, which one correctly illustrates how light rays come off an object? Explain your answer. 2. Of the diagrams below, which one correctly illustrates how a light ray enters and exits a piece of thick glass? Explain your answer.

6 Skill Sheet 7.2 Ray Diagrams 3. In your own words, explain what happens to light as it enters glass from the air. Why does this happen? Use the terms refraction and index of refraction in your answer. 4. Of the diagrams below, which one correctly illustrates how parallel light rays enter and exit a converging lens? Explain your answer. 5. Draw a diagram of a converging lens that has a focal point of 0 centimeters. In your diagram, show three parallel lines entering the lens and exiting the lens. Show the light rays passing through the focal point of the lens. Be detailed in your diagram and provide labels. 2

7 Skill Sheet 7.2 Ray Diagrams 2. How to draw ray diagrams A ray diagram helps you see where the image produced by a lens appears. The components of the diagram include the lens, the principal axis, the focal point, the object, and three lines drawn from the tip of the object and through the lens. These light rays meet at a point and intersect on the other side of the lens. Where the light rays meet indicates where the image of the object appears. Example: A lens has a focal length of 2 centimeters. An object is placed 4 centimeters to the left of the lens. Follow the steps to make a ray diagram using this information. Trace the rays and predict where the image will form. Steps: Draw a lens and show the principal axis. Draw a line that shows the plane of the lens. Make a dot at the focal point of the lens on the right and left sides of the lens. Place an arrow (pointing upward and perpendicular to the principle axis) at 4 centimeters on the left side of the lens. Line : Draw a line from the tip of the arrow that is parallel to the principal axis on the left, and that goes through the focal point on the right of the lens. Line 2: Draw a line from the tip of the arrow that goes through the center of the lens (where the plane and the principal axis cross). Line 3: Draw a line from the tip of the arrow that goes through the focal point on the left side of the lens, through the lens, and parallel to the principal axis on the right side of the lens. Lines, 2, and 3 converge on the right side of the lens where the tip of the image of the arrow appears. The image is upside down compared with the object. 3

8 Skill Sheet 7.2 Ray Diagrams 3. Drawing ray diagrams. A lens has a focal length of 4 centimeters. An object is placed 8 centimeters to the left of the lens. Trace the rays and predict where the image will form. Is the image bigger, smaller, or inverted as compared with the object? 2. Challenge question: An arrow is placed at 3 centimeters to the left of a converging lens. The image appears at 3 centimeters to the right of the lens. What is the focal length of this lens? (HINT: Place a dot to the right of the lens where the image of the tip of the arrow will appear. You will only be able to draw lines and 2. Where does line cross the principal axis if the image appears at 3 centimeters?) 3. What happens when an object is placed at a distance from the lens that is less than the focal length? Use the term virtual image in your answer. 4

9 Name: Skill Sheet 7.3 Thin Lens Formula Here you will become familiar and practice with a mathematical formula called the thin lens formula. This formula gives scientists a way to calculate the location and the size of an image that is produced by a lens.. What is the thin lens formula? When you use the thin lens formula, you assume that the thickness of the lens is very small compared with the distance between the lens and the object or the image. The formula applies both to convex or converging lenses and concave or diverging lenses. Converging lenses are thicker in the center than in the edges. Diverging lenses are thinner in the center than in the edges. The thin lens formula is: Some important rules in using the thin lens formula: Object distance, d 0, is positive to the left of the lens and negative to the right of the lens. Image distance, d i, is negative to the left of the lens and positive to the right of the lens. Positive d 0, d i indicates real object or image. Negative d 0, d i indicates virtual object or image.

10 Skill Sheet 7.3 Thin Lens Formula 2. Examples in using the thin lens formula If you are using a convex lens, what happens an image when the object is very far from the lens? In other words, what happens when d 0 is a large number? Graphically, we see that as d 0 increases, the image becomes smaller as it gets closer to the focal point. We can see this by using the thin lens formula with a lens that has a focal length equal to 5 centimeters. If d 0 = then... d i = 8 cm, 00 cm,,000 cm, --- d i --- d i --- d i = = = = = = = = = , = 3.3 cm = 5.26 cm = 5.03 cm 0.99 If you are using a concave lens, you use a negative value for the focal length, f. In this case, the resulting image distance is a negative number indicating a virtual image. The calculations are shown below: If d 0 = then... d i = 8 cm, 00 cm,,000 cm --- d i --- d i --- d i = = = = = = = = = , = 4.7 cm = 4.76 cm = 4.98 cm

11 Skill Sheet 7.3 Thin Lens Formula 3. Problems. Calculate the location of the image if the object is 20 centimeters in front of a convex (converging) lens with a focal length of 5 centimeters. 2. The image of an object as seen by a converging lens is located at 8 centimeters. The object is 24 centimeters in front of the lens. What is the focal length? 3. For a converging lens with a focal length of 0 centimeters, calculate the location of the object when the image appears 20 centimeters to the right. 4. An object is 0 centimeters in front of a convex (converging) lens with a focal length of 5 centimeters. a. Calculate the location of the image. b. What kind of image appears in this situation? Where does it appear in relation to the lens? 5. Calculate the location of the image if the object is 20 centimeters in front of a concave (diverging) lens with a focal length of 5 centimeters. 6. An object is at a distance of 3 centimeters from a lens with a focal length of centimeter. The lens creates an image on the same side of the object. What kind of lens is this? What is the image location? Is the image real or virtual? 3

12 Skill Sheet 7.3 Thin Lens Formula 4. Problems that involve the height of images and objects Sample problem: The object is 0 centimeters in height and is located at a distance of 25 centimeters from the lens. The focal length is 8 centimeters. Find the image location d i and the height h of the image. Solution: Since we know the height of the object and the object distance d 0, we can calculate the angle α. α = tan = tan ( 0.4) = 2.8 The image distance d i can be found from the thin lens formula, and it is.7 centimeters. Now, since we know the image distance and the angle α, we can calculate the height. h = tanαd i = tan2.8.7 cm = 4.7 cm So we see that the image is actually smaller than the object.. Prove that the d i in the problem above is.7 centimeters by using the thin lens formula. 2. A 4-centimeter tall object seen through a lens with a focal length of 0 centimeters has an image half its size. a. Calculate the location of the image and the object. (HINT: Use proportions.) b. Using your answers for 2(a) and the height of the object, find the angle α. 4