PHY208FALL2008. Week3HW. Is Light Reflected or Refracted? Due at 11:59pm on Sunday, September 21, [ Print ] View Grading Details

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1 Assignment Display Mode: View Printable Answers PHY208FALL2008 Week3HW [ Print ] Due at 11:59pm on Sunday, September 21, 2008 View Grading Details The next exercise is about reflection and refraction of light Is Light Reflected or Refracted? Description: Mostly conceptual questions on index of refraction and Snell's law. Last few parts deal with total internal reflection. When light propagates through two adjacent materials that have different optical properties, some interesting phenomena occur at the interface separating the two materials. For example, consider a ray of light that travels from air into the water of a lake. As the ray strikes the air-water interface (the surface of the lake), it is partly reflected back into the air and partly refracted or transmitted into the water. This explains why on the surface of a lake sometimes you see the reflection of the surrounding landscape and other times the underwater vegetation. These effects on light propagation occur because light travels at different speeds depending on the medium. The index of refraction of a material, denoted by, gives an indication of the speed of light in the material. It is defined as the ratio of the speed of light in vacuum to the speed in the material, or. When light propagates from a material with a given index of refraction into a material with a smaller index of refraction, the speed of the light Hint A.1 Index of refraction The index of refraction of a material is defined as the ratio of the speed of light in vacuum to the speed in that particular material, or. Since it is the ratio of two positive quantities that have the same units, the index of refraction is a pure (positive) number. Note that the speed of light in a certain material is inversely proportional to the index of refraction of that material. increases. Part B What is the minimum value that the index of refraction can have? Hint B.1 Remember that the speed of light in a certain material is inversely proportional to the index of refraction of that material. Thus, the minimum value of the index of refraction is calculated for the medium where the speed of light is maximum. That occurs in vacuum where. between 0 and 1 The index of refraction of a material is always a positive number greater than 1 that tells us how fast the light travels in the material. The greater the index of refraction of a material, the more slowly light travels in the material. An example of reflection and refraction of light is shown in the figure. An incident ray of light traveling in the upper material strikes the interface with the lower material. The reflected ray travels back in the upper material, while the refracted ray passes into the lower material. Experimental studies have shown that the incident, reflected, and refracted rays and the normal to the interface all lie in the same plane. Moreover, the angle that the Page 1 of 13

2 reflected ray makes with the normal to the interface, called the angle of reflection, is always equal to the angle of incidence. (Both of these angles are measured between the light ray and the normal to the interface separating the two materials.) This is known as the law of reflection. The direction of propagation of the refracted ray, instead, is given by the angle that the refracted ray makes with the normal to the interface, which is called the angle of refraction. The angle of refraction depends on the angle of incidence and the indices of refraction of the two materials. In particular, if we let be the index of refraction of the upper material and the index of refraction of the lower material, then the angle of incidence,, and the angle of refraction,, satisfy the relation. This is the law of refraction, also known as Snell's law. Part C Now consider a ray of light that propagates from water ( ) to air ( ). If the incident ray strikes the water-air interface at an angle, which of the following relations regarding the angle of refraction,, is correct? Part C.1 Find an expression for the ratio of the sines of and Let the index of refraction of water be and that of air be. Use Snell's law to find an expression for the ratio of the sine of to the sine of. Express your answer in terms of some or all of the variables,, and. Now, note that for the water-air interface. Therefore,. When light propagates from a certain material to another one that has a smaller index of refraction, that is, speed of propagation of the light rays increases and the angle of refraction is always greater than the angle of incidence. This means that the rays are always bent away from the normal to the interface separating the two media., the Part D Consider a ray of light that propagates from water ( ) to glass ( ). If the incident ray strikes the water-glass interface at an angle, which of the following relations regarding the angle of refraction is correct? Part D.1 Find an expression for the ratio of the sines of and Let the index of refraction of water be and that of glass be. Use Snell's law to find an expression for the ratio of the sine of to the sine of. Express your answer in terms of some or all of the variables,, and. Now, note that for the water-glass interface. Therefore,. When light propagates from a certain material to another one that has a greater index of refraction, that is, speed of propagation of the light rays decreases and the angle of refraction is always smaller than the angle of incidence. This means that the rays are always bent toward the normal to the interface separating the two media., the Page 2 of 13

3 Part E Consider a ray of light that propagates from air ( the interface with any of the listed materials always at the same angle the ray change the most due to refraction? Hint E.1 ) to any one of the materials listed below. Assuming that the ray strikes, in which material will the direction of propagation of The direction of propagation of the ray of light will change the most when the difference between the angle of refraction and the angle of incidence is maximum. Since we are studying a situation where light propagates from air to a material that has a greater index of refraction, we can make use of the results obtained in Part D. We know that, in this case, the angle of refraction is always smaller than the angle of incidence. Thus, the difference between the angle of refraction and the angle of incidence is maximum when the angle of refraction is smallest. Part E.2 Find an expression for the sine of the angle of refraction Let the index of refraction of the unknown material be refraction,. Express your answer in terms of some or all of the variables and.. Use Snell's law to find an expression for the sine of the angle of Your result shows that the sine of the angle of refraction is inversely proportional to the index of refraction of the unknown material. Therefore, the angle of refraction is minimum in the material that has the greatest index of refraction. ice ( ) water ( ) turpentine ( ) glass ( ) diamond ( ) The greater the change in index of refraction, the greater the change in the direction of propagation of light. To avoid or minimize undesired bending of the light rays, light should travel through materials with matching indices of refraction. Is light always both reflected and refracted at the interface separating two different materials? To answer this question, let's consider the case of light propagating from a certain material to another material with a smaller index of refraction (i.e., ). Part F In the case of, if the incidence angle is increased, the angle of refraction Hint F.1 How to approach the question Recall that, according to Snell's law, the sine of the angle of refraction is directly proportional to the sine of the angle of incidence. Thus, as the angle of incidence is increased, the angle of refraction changes accordingly. Moreover, since the angle of refraction is greater than the angle of incidence, as you found in Part C, the angle of refraction can reach its maximum value sooner than the angle of incidence. increases up to a maximum value of 90 degrees. Since the light is propagating into a material with a smaller index of refraction, the angle of refraction,, is always greater than the angle of incidence,. Therefore, as is increased, at some point will reach its maximum value of 90 and the refracted ray will travel along the interface. The angle of incidence for which is called the critical angle. For any angle of incidence greater than, no refraction occurs. The ray no longer passes into the second material. Instead, it is completely reflected back into the original material. This phenomenon is called total internal reflection and occurs only when light encounters an interface with a second material with a smaller index of refraction than the original material. Part G What is the critical angle refraction of 1.00? for light propagating from a material with index of refraction of 1.50 to a material with index of Part G.1 Find an expression for the sine of the angle of incidence Use Snell's law to find a general expression for the sine of the angle of incidence, material with index of refraction to a material with index of refraction., for a ray of light that travels from a Express your answer in terms of,, and. Page 3 of 13

4 Now find when, 1.50, and Express your answer in radians. In conclusion, light is always both reflected and refracted, except in the special situation when the conditions for total internal reflection occur. In that case, there is no refracted ray and the incident ray is completely reflected. The next two problems are applications of snell's law of refraction Problem Description: An underwater diver sees the sun 50 degree(s) above horizontal. (a) How high is the sun above the horizon to a fisherman in a boat above the diver? An underwater diver sees the sun 50 above horizontal. How high is the sun above the horizon to a fisherman in a boat above the diver? Problem Description: The glass core of an optical fiber has an index of refraction The index of refraction of the cladding is (a) What is the maximum angle a light ray can make with the wall of the core if it is to remain inside the fiber? The glass core of an optical fiber has an index of refraction The index of refraction of the cladding is What is the maximum angle a light ray can make with the wall of the core if it is to remain inside the fiber? Underwater Optics Description: The net refraction of light passing through several materials is calculated and then used to find the apparent distance of objects underwater. Your eye is designed to work in air. Surrounding it with water impairs its ability to form images. Consequently, scuba divers wear masks to allow them to form images properly underwater. However, this does affect the perception of distance, as you will calculate. Consider a flat piece of plastic (index of refraction ) with water (index of refraction ) on one side and air (index of refraction ) on the other. If light is to move from the water into the air, it will be refracted twice: once at the water/plastic interface and once at the plastic/air interface. If the light strikes the plastic (from the water) at an angle, at what angle does it emerge from the plastic (into the air)? Hint A.1 Angles inside the plastic There are two important angles within the plastic: the angle immediately after the first refraction (the water/plastic interface) and the angle immediately before the second refraction (the plastic/air interface). To find out how they relate, draw a picture with the path the light follows in the plastic and the normals to both surfaces. Once you have labeled both angles, keep in mind that the surfaces are parallel, and thus their normals are parallel lines. An important theorem from geometry will give you the relationship between the angles. Hint A.2 Important theorem from geometry If two parallel lines are cut by a transversal (a third line not parallel to the first two), then alternate interior angles are congruent. Page 4 of 13

5 .3 Find the angle in the plastic Using Snell's law, find, the angle inside the plastic (i.e., the angle to the normal immediately after the first refraction at the water/plastic interface). Hint A.3.a Snell's law Snell's law states that, where is the angle of incidence, is the index of refraction for the medium from which light is incident, is the angle of refraction, and is the index of refraction for the medium into which light emerges. Express your answer in terms of,, and. Remember that the inverse sine of a number should be entered as asin(x) in the answer box. Now, when you calculate, don't forget that. Express your answer in terms of,,, and. Remember that the inverse sine of a number should be entered as asin(x) in the answer box. Notice that does not appear in this equation. Humans estimate distance based on several different factors, such as shadows and relative positions. The most important method for estimating distance, triangulation, is performed unconsciously. Triangulation is based on the fact that light from distant objects strikes each eye at a slightly different angle. Your brain can then use that information to determine the angle as shown in the figure. In the figure, points L and R represent your left and right eyes, respectively. The distance between your eyes is, and the distance to the object, point O, is. Part B What is the distance to the object in terms of and? Express your answer in terms of and. Part C If the distance to the object is more than about 0.4, then you can use the small-angle approximation. What is the formula for the distance to the object, if you make use of this approximation? Express your answer in terms of and. Page 5 of 13

6 Your eyes determine by assuming that and (in the figure) are equal. This is true, unless the light rays are bent before they reach your eyes, as they are if you're wearing a scuba mask underwater. Underwater, the situation changes, as shown in the figure..your eyes will calculate an apparent distance using the angle that reaches your eyes, instead of the correct geometric angle is the same. This that you calculated in. Note that there are no important geometric considerations arising from the refraction except the substitution of for, because the refraction takes place so close to your eyes. If the problem discussed someone looking out of the porthole of a submarine, the geometry would become more complicated. Part D Now use the expression found in Part C for the distance between your eyes and the object at point O, and find the ratio of the apparent distance to the real distance,. Remember that the apparent distance is the distance calculated by your eyes using the angle instead of the angle. Since we are dealing with small angles, you may use the approximations and. Part D.1 Use the small-angle approximations From, you have the expression. Apply both small-angle approximations to this equation to get a simpler expression for. Express your answer in terms of,, and. Part D.2 Find Because of the refraction your eyes use instead of. Once you've used the small-angle approximations, plug your equation for into the equation from Part C,. Since you are putting into the equation instead of, this gives the apparent distance. Express your answer in terms of,,, and. Express your answer in terms of and. Part E Given that and, by what percent do objects underwater appear closer than they actually are? Hint E.1 To find the fraction by which objects appear closer, simply find the difference between the apparent distance and the true distance, then divide by the true distance. You can then convert this fraction to a percent to get the final answer. For example, if an object is 1 away and it appears to be 0.6 away then it is 40% closer ( %). Page 6 of 13

7 Express your answer to two significant figures. % These problems deal with ray optics and lenses. Problem 23.5 Description: A student has built a l-long pinhole camera for a science fair project. She wants to photograph the Washington Monument, which is 167 m (550 ft) tall, and to have the image on the film be h high. (a) How far should she stand from the Washington... A student has built a long pinhole camera for a science fair project. She wants to photograph the Washington Monument, which is 167 m (550 ft) tall, and to have the image on the film be 5.10 How far should she stand from the Washington Monument? high. m A Laser and a Lens Description: A laser is projected through a lens onto a screen. Find the location of the point on the screen. A laser is mounted as shown in the figure (distances and are given) above the axis of a converging lens with positive focal length. The laser beam travels parallel to the axis of the lens. A large screen is placed at a distance The laser beam passes through the lens and makes a dot on the screen at a distance to the right of the lens., measured upward from the axis of the lens. Assume that a positive value means that the dot is above the axis, while a negative value means that the dot is below the axis. Find the distance. Hint A.1 Make your own diagram and draw the refracted ray. Then use the geometry of similar triangles to find the position of the dot on the screen. Hint A.2 Drawing the refracted ray Since the incident ray is parallel to the axis of the lens, the refracted ray passes through the focal point on the right side of the lens. Page 7 of 13

8 .3 Find a pair of similar triangles Once you have drawn your own diagram with the refracted ray shown, similar to the picture shown here, consider the two similar triangles ACF and EFG that the refracted ray forms with the axis of the lens, one on each side of the focal point F. The triangle ACF has sides of length and. What are the lengths of the corresponding sides of the triangle EFG? and and and and Now use triangle similarity to find. Note that your final answer should refer to the location of the dot on the screen. If the dot on the screen is above the axis of the lens, the corresponding length should be positive. If the dot on the screen is below the axis of the lens, the corresponding length should be negative. Express your answer in terms of some or all of the variables,,, and. Note that not all of the variables may be required for the answer. Notice that the sign of the height depends upon the sign of. If the focal point is in front of the screen, the dot will be below the axis of the lens. If the focal point is on the screen, the image will be at the level of the axis of the lens, and if the focal point is behind the screen, the dot will be above the axis of the lens. Now consider a specific case. Let the laser be 50 centimeters to the left of the lens and at height 15 centimeters above the axis of the lens. The lens has focal length 30 centimeters, and the screen is 1 meter to the right of the lens. Part B What is the position of the dot on the screen? Express your answer in centimeters, to two significant figures. Part C If the laser is moved farther from the lens, what happens to the dot on the screen? Hint C.1 Consider the expression found in. Does the position of the dot on the screen depend on the distance of the laser from the lens? It moves up. It moves down. It does not move. Part D If the laser is moved up, farther above the axis of the lens, what happens to the dot on the screen? Assume that the ray still strikes the lens. Hint D.1 Page 8 of 13

9 Recall the expression found in. If you double the value of, how is affected? It moves up. It moves down. It does not move. Focusing with the Human Eye Description: A man looks at three objects at different distances. Explain how he must adjust the lenses in his eyes to bring the various objects into focus. Joe is hiking through the woods when he decides to stop and take in the view. He is particularly interested in three objects: a squirrel sitting on a rock next to him, a tree a few meters away, and a distant mountain. As Joe is taking in the view, he thinks back to what he learned in his physics class about how the human eye works. Light enters the eye at the curved front surface of the cornea, passes through the lens, and then strikes the retina and fovea on the back of the eye. The cornea and lens together form a compond lens system. The large difference between the index of refraction of air and that of the aqueous humor behind the cornea is responsible for most of the bending of the light rays that enter the eye, but it is the lens that allows our eyes to focus. The ciliary muscles surrounding the lens can be expanded and contracted to change the curvature of the lens, which in turn changes the effective focal length of the cornea-lens system. This in turn changes the location of the image of any object in one's field of view. Images formed on the fovea appear in focus. Images formed between the lens and the fovea appear blurry, as do images formed behind the fovea. Therefore, to focus on some object, you adjust your ciliary muscles until the image of that object is located on the fovea. Joe first focuses his attention (and his eyes) on the tree. The focal length of the cornea-lens system in his eye must be the distance between the front and back of his eye. Hint A.1 Draw a picture of the object (the tree), the lens, and the image it produces. Be sure to include the focal point of the lens. Where must the fovea be in your sketch if this object is in focus? Is the focal point between the lens and the fovea, on the fovea, or behind the fovea? greater than less than equal to Part B Joe's eyes are focused on the tree, so the squirrel and the mountain appear out of focus. This is because the image of the squirrel is formed and the image of the mountain is formed. Hint B.1 Image of the squirrel The squirrel is closer to the lens (the eye) than the tree. As long as Joe's eyes stay fixed on the tree, their focal length does not change. Using the lens equation, determine whether the image of the squirrel is closer to the lens than the image of the tree or farther away. Hint B.2 Image of the mountain The mountain is farther from the lens (the eye) than the tree. As long as Joe's eyes stay fixed on the tree, their focal length does not change. Using the lens equation, determine whether the image of the mountain is closer to the lens than the image of the tree or farther away. between the lens and fovea / between the lens and fovea between the lens and fovea / behind the fovea behind the fovea / between the lens and fovea behind the fovea / behind the fovea Part C Joe now shifts his focus from the tree to the squirrel. To do this, the ciliary muscles in his eyes must have the curvature of the lens, resulting in a(n) focal length for the cornea-lens system. Note that curvature is different from radius of curvature. Page 9 of 13

10 increased / increased increased / decreased decreased / increased decreased / decreased Part D Finally, Joe turns his attention to the mountain in the distance but finds that he cannot bring the mountain into focus. This is because he is nearsighted. But when Joe puts on his glasses, he can see the mountain clearly. To adjust for his nearsightedness, his glasses must contain lenses. Hint D.1 Focusing on distant objects The image of a distant object like the mountain always forms at (or very close to) the focal point of the fovea-lens system. When Joe is looking at the most distant object he can see clearly, where is the focal point? Hint D.2 The role of corrective lenses Nearsightedness and farsightedness are both caused by the fact that the ciliary muscles cannot make the focal length of the lens arbitrarily large or small. The corrective lenses must make the image of the distant mountains form someplace that his eyes are naturally able to focus on. converging diverging A Two-Lens System Description: Calculate the image size and position for a two-lens system. Then, find the lens that would put the image in the same position if it replaced the two lenses. A compound lens system consists of two converging lenses, one at with focal length, and the other at with focal length. An object centimeter tall is placed at. What is the location of the final image produced by the compound lens system? Give the x coordinate of the image. Hint A.1 How to handle multiple optics The image formed by the first lens acts as the object for the second lens..2 Find the object distance for the first lens How far is the object from the first lens?.3 Find the image distance from the first lens Ignoring the second lens, determine where the image is formed just by the first lens. Give its distance from the lens..4 Find the object distance for the second lens Page 10 of 13

11 How far is the image produced by the first lens from the second lens?.5 Find the image distance from the second lens Using the result of the previous hint, determine how far the final image is from the second lens. Part B How tall is the image? Hint B.1 The total magnification is the product of the magnifications caused by the two lenses seperately:. If you have difficulty finding the individual magnifications, use the other hints. Part B.2 Find the magnification of the first lens What is the magnification of the first lens? Recall that magnification is defined in two ways: and. Express your answer to three significant figures or as a fraction. Part B.3 Find the magnification of the second lens What is the magnification of the second lens? Recall that magnification is defined in two ways: and. Express your answer to three significant figures or as a fraction. Part C Is the final image upright or inverted, relative to the original object at? upright inverted Now remove the two lenses at and and replace them with a single lens of focal length at. We want to choose this new lens so that it produces an image at the same location as before. Part D What is the focal length of the new lens at the origin? Part D.1 Find the object distance for the third lens How far is the object from the new lens? Page 11 of 13

12 Part D.2 Find the image distance for the third lens How far is the image from the new lens? Part E Is the image formed by orientation? the same size as the image formed by the compound lens system? Does it have the same Part E.1 Find the magnification of the third lens What is the magnification of the third lens? Compare the result with your answer for Parts B and C. Express your answer to three significant figures or as a fraction. The image is the same size and oriented the same. The image is the same size and oriented differently. The image is a different size and oriented the same. The image is a different size and oriented differently. A Microscope for Biology Description: Find the length of a microscope tube needed to view an object in focus, given its distance from the objective and the objective focal length. In a two-lens system, the image produced by one lens acts as the object for the next lens. This simple principle finds applications in many optical instruments, including some of common use such as the microscope and the telescope. The microscope available in your biology lab has a converging lens (the eyepiece) with a focal length of 2.50 mounted on one end of a tube of adjustable length. At the other end is another converging lens (the objective) that has a focal length of When you place the sample to be examined at a distance of 1.30 from the objective, at what length will you need to adjust the tube of the microscope in order to view the sample in focus with a completely relaxed eye? Note that to view the sample with a completely relaxed eye, the eyepiece must form its image at infinity. Hint A.1 Since the microscope is a two-lens system, there are two steps to follow to solve the problem. First calculate the location of the image formed by the objective. Then consider this image to be the object for the eyepiece and find its distance from the eyepiece. The length of the tube will be the sum of the image distance for the objective and the object distance for the eyepiece..2 Find the image distance for the objective What is the image distance for the objective, given that its focal length is 1.00 and the object distance is 1.30? Hint A.2.a The thin-lens equation The image distance can be found from the thin-lens equation, where is the object distance and is the focal length of the lens. Page 12 of 13

13 Express your answer in centimeters..3 Find the object distance for the eyepiece What is the object distance for the eyepiece, given that it forms an image at infinity and its focal length is 2.50?.3.a Find where the object is located At what point should the object for the eyepiece be located if the eyepiece is to form an image at infinity? at the center of curvature of the eyepiece at the focal point of the eyepiece at a point between the focal point and the center of curvature of the eyepiece at infinity at the focal point of the objective Express your answer in centimeters. The length of the tube is then the sum of the image distance for the objective and the object distance for the eyepiece. Express your answer in centimeters. Problem Description: A slide projector needs to create a h-high image of a h1-tall slide. The screen is l from the slide. (a) What focal length does the lens need? Assume that it is a thin lens. (b) How far should you place the lens from the slide? A slide projector needs to create a high image of a tall slide. The screen is 294 from the slide. What focal length does the lens need? Assume that it is a thin lens. cm Part B How far should you place the lens from the slide? cm Summary 0 of 10 items complete (0% avg. score) 0 of 50 points Page 13 of 13

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