HW 10 Ch 24 Optical InstrumentsDue: 11:59pm on Tuesday, November 24, 2015 To understand how points are awarded, read the Grading Policy for this assignment. A TwoLens System A compound lens system consists of two converging lenses, one at x = −20.0 cm with focal length f 1 = +10.0 cm, and the other at x = +20.0 cm with focal length f 2 = +8.00 cm. An object 1.00 centimeter tall is placed at x = −50.0 cm . Part A What is the location of the final image produced by the compound lens system? Give the x coordinate of the image. Express your answer in centimeters, to three significant figures or as a fraction. Hint 1. How to handle multiple optics The image formed by the first lens acts as the object for the second lens. Hint 2. Find the object distance for the first lens How far is the object from the first lens? Express your answer in centimeters, to three significant figures or as a fraction. ANSWER: s ob = cm Hint 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. Express your answer in centimeters, to three significant figures or as a fraction. ANSWER: s im = cm Hint 4. Find the object distance for the second lens How far is the image produced by the first lens from the second lens? Express your answer in centimeters, to three significant figures or as a fraction. ANSWER: s ob = cm Hint 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. Express your answer in centimeters, to three significant figures or as a fraction. ANSWER: s im = cm ANSWER: x = 31.8 cm Correct Part B How tall is the image? Express your answer in centimeters, to three significant figures or as a fraction. Hint 1. How to approach the problem The total magnification m is the product of the magnifications caused by the two lenses seperately: m = m 1 m 2 . If you have difficulty finding the individual magnifications, use the other hints. Hint 2. Find the magnification of the first lens What is the magnification of the first lens? Recall that magnification is defined in two ways: m = m=− s im s ob . Express your answer to three significant figures or as a fraction. y im y ob and ANSWER: m1 = Hint 3. Find the magnification of the second lens What is the magnification of the second lens? Recall that magnification is defined in two ways: m = and m = − s im s ob y im y ob . Express your answer to three significant figures or as a fraction. ANSWER: m2 = ANSWER: y im = 0.236 cm Correct Part C Is the final image upright or inverted, relative to the original object at x = −50 cm ? ANSWER: upright inverted Correct Now remove the two lenses at x = +20.0 cm and x = −20.0 cm and replace them with a single lens of focal length f at x = 0. We want to choose this new lens so that it produces an image at the same location as before. 3 Part D What is the focal length of the new lens at the origin? Express your answer in centimeters, to three significant figures or as a fraction. Hint 1. Find the object distance for the third lens How far is the object from the new lens? Express your answer in centimeters, to three significant figures or as a fraction. ANSWER: s ob = cm Hint 2. Find the image distance for the third lens How far is the image from the new lens? Express your answer in centimeters, to three significant figures or as a fraction. ANSWER: s im = cm ANSWER: f 3 = 19.4 cm Correct Part E Is the image formed by f 3 the same size as the image formed by the compound lens system? Does it have the same orientation? Hint 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. ANSWER: m3 = ANSWER: 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. Correct ± Understanding Multiple Optics Learning Goal: To become familiar with using the image of one instrument as the object of the next and tracing rays through a system of multiple instruments. Multiple optics refers to any system of more than one optical instrument through which light passes. Most devices related to optics, such as cameras, microscopes, and telescopes, contain multiple optics systems. In multiple optics, the image of one optical instrument becomes the object of the next one. Thus, in multiple optics problems, you need to find the image created by the first optical instrument that the rays encounter. Then, you will use that image as the object of the next optical instrument, repeating this pattern until you have followed the rays all the way through the system. It is very important to be alert to the geometry and to signs when you find the object distance for one instrument from the location of the previous instrument's image. Sometimes, the image is formed on the virtual side of the instrument, leading to a virtual object. This may sound strange, but in practice, its effect on your calculations is simply to make the object distance negative instead of positive. Several optical instruments are placed along the x axis, with their axes aligned along the x axis. A plane mirror is located at x = 0. A converging lens with focal length 5.00 m is located at x = 12.5 m . An object is placed at x = 22.5 m . In order to find the location of the final image of the object formed by this system, you will need to trace the rays through the system, instrument by instrument. You are strongly advised to draw a picture with the x axis and the location of the lens, mirror, and object marked. Then, as you proceed through the problem, you can mark where each image is located. Part A First, find the location of the image created by the lens by itself (as if no other instruments were present). Express your answer in meters, to three significant figures, or as a fraction. Hint 1. The thin lens equation For a thin lens, one has the relation 1 s + 1 s′ = 1 f , where s ′ is the image distance, s is the object distance, and f is the focal length. ANSWER: x = 2.50 m Correct Part B Next, find the location of the image created by the plane mirror (after the light has passed through the lens). Express your answer in meters, to three significant figures, or as a fraction. Hint 1. The focal length of a plane mirror Keep in mind that a plane mirror has a focal length of infinity (since a plane mirror has an infinite radius of curvature). Hint 2. Find the object distance You know from Part A that the image resulting from the passage of light though the lens is located at x = 2.50 m . If the plane mirror is at x = 0, what is the object distance s for the plane mirror? Express your answer in meters to three significant figures. ANSWER: s = m ANSWER: x = 2.50 m Correct Proceed with caution! The light reflects off of the mirror and back through the lens a second time! Part C What is the location of the final image, as seen by an observer looking toward the mirror, through the lens? Keep in mind that the light must pass back through the lens, and thus you must do one more calculation with the thin lens equation. Express your answer in meters, to three significant figures, or as a fraction. Hint 1. Object distance for the lens (second time) What is the object distance s for the lens the second time? Recall that the object is the image made by the mirror. You found its coordinate in the last part, so just subtract that from the coordinate of the lens to find the object distance for the lens the second time. Express your answer in meters to three significant figures or as a fraction. ANSWER: s = ANSWER: m x = 20.0 m Correct Part D Is the final image formed by this system real or virtual? Hint 1. Real versus virtual A real image is one that forms at a location that the light actually reaches (imagine a slide projector creating an image on a screen). A virtual image is one that forms at a location that the light does not actually reach. (Imagine the image created when you look in a mirror hanging on a wall. No light actually makes it past the wall; there only appears to be a copy of you on the other side of the wall!) Hint 2. Real versus virtual with multiple optics Since the type of image is determined only by whether the light actually reaches the location of the image, whether the image is real or virtual depends only on the last optical instrument through which the light passes. Thus, the light may fail to reach an intermediate image (in this example, the one created by the mirror, which was a virtual image) but may still create a real final image, which it will do if the light reaches the location of the final image created by the system. ANSWER: real virtual Correct If you have more than one optical instrument, the total magnification is equal to the product of the individual magnifications. Keep in mind that the image of one device becomes the object of the next. If the first device creates an image that is ten times the size of the object, and the second creates an image that is twenty times the size of the first image (which was its object), then the final image will be two hundred times the size of the original object. This makes sense because the second device magnifies further what was already magnified by the first device. To find the magnitude of the magnification of the final image, you will need to consider each instrument and find the individual magnifications. Part E First, find the magnitude mlens1 of the magnification of the image created when light from the object passes through the lens the first time (as if the mirror were not present). Express your answer to three significant figures or as a fraction. Hint 1. Magnification The magnification of an image is defined as the negative of the ratio of the image distance to the object distance. The magnitude determines the relative size of the image as compared to the object, while the sign yields the orientation (negative is inverted, positive is upright). ANSWER: m lens1 = 1 Correct Part F Next, find the magnitude mmirror of the magnification of the plane mirror. Express your answer to three significant figures or as a fraction. ANSWER: m mirror = 1 Correct Of course, this is expected. Plane mirrors don't magnify your image! Part G Now find the magnitude mlens2 of the magnification of the image created when light from the object passes through the lens the second time (after reflecting off the mirror). Express your answer to three significant figures or as a fraction. ANSWER: m lens2 = 0.5000 Correct Part H What is the magnitude of the magnification of the final image? Express your answer to three significant figures or as a fraction. ANSWER: m = 0.5000 Correct Part I Is the final image upright or inverted? Give the orientation relative to the original object. Hint 1. Magnification in multiple optics (orientation) If you have more than one optical instrument, the total magnification is just the product of the individual magnifications. The image of one device becomes the object of the next. If, for example, the first device creates an image that is inverted (negative magnification) and the second creates one that is also inverted (negative magnification), then the final image will be upright (positive magnification). Mathematically, this makes sense because the product of two negative numbers is positive. Conceptually, this makes sense because if the first device inverts the image, and the second also inverts the image, then the first device flips the image upside down, and the second flips it back again, resulting in an image that is rightside up. Hint 2. Find the orientation from the lens (first time) Is the image created by the lens (as if the mirror were not present) inverted or upright? ANSWER: upright inverted Hint 3. Find the orientation from the mirror Is the image created by the mirror (of the image created by the lens) inverted or upright, compared to the image created by the lens? ANSWER: upright inverted Hint 4. Find the orientation from the lens (second time) Is the image created by the lens (of the image created by the mirror) inverted or upright, compared to the image created by the mirror? ANSWER: upright inverted ANSWER: upright inverted Correct Corrective Lenses Conceptual Questions The near point (the smallest distance at which an object can be seen clearly) and the far point (the largest distance at which an object can be seen clearly) are measured for six different people. near point far point (cm) (cm) Avishka 40 ∞ Berenice 30 300 Chadwick 25 500 Danya 25 ∞ Edouard 80 200 Francesca 50 ∞ Part A Which, if any, of these people are nearsighted (myopic)? List the first letter of all correct answers in alphabetical order. For example, if Avishka and Edouard are the only nearsighted ones, enter AE. Hint 1. Being nearsighted (myopic) A person with normal vision can focus clearly on objects an infinite distance away. A person is considered nearsighted, or myopic, if their fully relaxed eye focuses only out to a finite distance. This is the result of the eye converging the light from infinity in too short a distance. ANSWER: BCE Correct Nearsightedness can be corrected by diverging lenses, which form virtual images at the individual's far point when presented with objects at infinity. Part B Which, if any, of these people require bifocals to correct their vision? List the first letter all correct answers in alphabetical order. For example, if Avishka and Edouard are the ones needing bifocals, enter AE. Hint 1. Being farsighted (hyperopic) Those with normal vision can focus on objects at a normal reading distance, defined to be 25 cm. A person is considered farsighted, or hyperopic, who is unable to focus on objects within this distance. This problem is the result of the eye not converging the light from a nearby object strongly enough. Hint 2. Bifocals A person who is both myopic and hyperopic will need bifocals to see both far and near objects clearly. Bifocals typically have a relatively small converging lens set in the bottom portion of an otherwise diverging lens. To see nearby objects, for example to read, the person looks through the bottom converging portion of the lens. ANSWER: BE Correct Farsightedness can be corrected by converging lenses, which form virtual images at the individual's near point when presented with objects at 25 cm. Bifocals include both converging and diverging lenses to correct for both farsightedness and nearsightedness. Part C Which, if any, of these people's vision can be corrected using only converging lenses? List all correct answers in alphabetical order. For example, if Avishka and Edouard are the ones whose vision can be thus corrected, enter AE. ANSWER: AF Correct Part D Of the farsighted people, rank them by the power of the lens needed to correct their hyperopic vision. Rank these from largest to smallest power required. To rank items as equivalent, overlap them. Hint 1. Power The power of a lens is the inverse of its focal length. Lenses with small focal lengths have large power and can correct more serious vision defects. Hint 2. Deviation from "normal" near point The near point of a normal eye is 25 cm. The larger the difference of a person's near point from this normal value, the more serious the vision defect. ANSWER: Correct Nearsightedness and Farsightedness A person with normal vision can focus on objects as close as a few centimeters from the eye up to objects infinitely far away. There exist, however, certain conditions under which the range of vision is not so extended. For example, a nearsighted person cannot focus on objects farther than a certain point (the far point), while a farsighted person cannot focus on objects closer than a certain point (the near point). Note that even though the presence of a near point is common to everyone, a farsighted person has a near point that is much farther from the eye than the near point of a person with normal vision. Both nearsightedness and farsightedness can be corrected with the use of glasses or contact lenses. In this case, the eye converges the light coming from the image formed by the corrective lens rather than from the object itself. Part A When glasses (or contact lenses) are used to correct nearsightedness, where should the corrective lens form an image of an object located at infinity in order for the eye to form a clear image of that object? Hint 1. Range of vision in nearsightedness To be effective, the corrective lens should form an image at a point located within the range of clear vision. Recall that a nearsighted person can form a clear image of any object located as far from the eye as the far point, but not farther. ANSWER: The lens should form the image at the near point. The lens should form the image at the far point. The lens should form the image at a point closer to the eye than the near point. The lens should form the image at a point farther from the eye than the far point. Correct This effect is achieved by the use of a diverging lens, as shown in the figure. Part B If a nearsighted person has a far point d f that is 3.50 m from the eye, what is the focal length f 1 of the contact lenses that the person would need to see an object at infinity clearly? Express your answer in meters. Hint 1. How to approach the problem Once you have determined the object distance and the image distance, you can use the thinlens equation to find the focal length of the contact lenses. Notice that the information found in Part A will help you find the image distance. Hint 2. Find the object distance At what distance d o from the contact lens is the object? Hint 1. Contact lenses Contact lenses are placed directly against the eye, so the lensobject distance is the same as the eyeobject distance. ANSWER: df do = −d f ∞ −∞ Hint 3. Find the image distance At what distance d i from the contact lens should the image form in order for the eye to focus clearly on it? Express your answer in meters. Hint 1. Contact lenses as corrective lenses As you found in Part A, a corrective lens should form the image at the far point. Also note that contact lenses are placed directly against the eye, so the eyeobject distance is the same as the lensobject distance. ANSWER: di = m ANSWER: f 1 = 3.50 m Correct Part C When glasses (or contact lenses) are used to correct farsightedness, where should the corrective lens form an image of an object located between the eye and the near point in order for the eye to form a clear image of that object? Hint 1. Range of vision in farsightedness To be effective, the corrective lens should form an image at a point located within the range of clear vision. Recall that a farsighted person can form a clear image of any object located as close to the eye as the near point, but not closer. ANSWER: The lens should form the image at the near point. The lens should form the image at the far point. The lens should form the image at a point closer to the eye than the near point. The lens should form the image at a point farther from the eye than the far point. Correct This effect is achieved by the use of a converging lens, as shown in the figure. Part D If a farsighted person has a near point that is 0.600 m from the eye, what is the focal length f 2 of the contact lenses that the person would need to be able to read a book held at 0.350 m from the person's eyes? Express your answer in meters. Hint 1. How to approach the problem Once you have determined the object distance and the image distance, you can use the thinlens equation to find the focal length of the contact lenses. Notice that the information found in Part C will help you find the image distance. Hint 2. Find the object distance At what distance d o from the contact lens is the object? Express your answer in meters. Hint 1. Contact lenses Contact lenses are placed directly against the eye, so the lensobject distance is the same as the eyeobject distance. ANSWER: do = m Hint 3. Find the image distance At what distance d i from the contact lens should the image form in order for the eye to focus clearly on it? Express your answer in meters. Hint 1. Corrective lenses Use the information you obtained in Part C. That is, keep in mind that the corrective lens should form the image at the near point and that contact lenses are placed directly against the eye. ANSWER: di = m ANSWER: f2 = 0.840 m Correct Problem 24.44 Your task in physics laboratory is to make a microscope from two lenses. One lens has a focal length of 2.2 cm , the other 1.1 cm . You plan to use the more powerful lens as the objective, and you want the eyepiece to be 15 cm from the objective. Part A For viewing with a relaxed eye, how far should the sample be from the objective lens? Express your answer to two significant figures and include the appropriate units. ANSWER: s = 1.08 cm All attempts used; correct answer withheld by instructor Part B What is the magnification of your microscope? Express your answer using two significant figures. ANSWER: M = 120 Correct Problem 24.15 A magnifier has a magnification of 9× . Part A How far from the lens should an object be held so that its image is seen at the nearpoint distance of 25 cm? Assume that your eye is immediately behind the lens. Express your answer to two significant figures and include the appropriate units. ANSWER: 2.8 cm Incorrect; Try Again; 3 attempts remaining Score Summary: Your score on this assignment is 47.0%. You received 2.35 out of a possible total of 5 points, plus 0 points of extra credit.