Concept explainers
A floating strawberry illusion is achieved with two parabolic mirrors, each having a focal length 7.50 cm, facing each other as shown in Figure P33.58. If a strawberry is placed on the lower mirror, an image of the strawberry is formed at the small opening at the center of the top mirror, 7.50 cm above the lowest point of the bottom mirror. The position of the eye in Figure P35.58a corresponds to the view of the apparatus in Figure P35.58b. Consider the light path marked A. Notice that this light path is blocked by the upper mirror so that the strawberry itself is not directly observable. The light path marked B corresponds to the eye viewing the image of the strawberry that is formed at the opening at the top of the apparatus. (a) Show that the final image is formed at that location and describe its characteristics. (b) A very startling effect is to shine a flashlight beam on this image. Even al a glancing angle, the incoming light beam is seemingly reflected from the image! Explain.
Figure P35.58
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Chapter 36 Solutions
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
- In Figure P35.30, a thin converging lens of focal length 14.0 cm forms an image of the square abed, which is he = hb = 10.0 cm high and lies between distances of pd = 20.0 cm and pa = 30.0 cm from the lens. Let a, b, c. and d represent the respective corners of the image. Let qa represent the image distance for points a and b, qd represent the image distance for points c and d, hb, represent the distance from point b to the axis, and hc represent the height of c. (a) Find qa, qd, hb, and hc. (b) Make a sketch of the image. (c) The area of the object is 100 cm2. By carrying out the following steps, you will evaluate the area of the image. Let q represent the image distance of any point between a and d, for which the object distance is p. Let h represent the distance from the axis to the point at the edge of the image between b and c at image distance q. Demonstrate that h=10.0q(114.01q) where h and q are in centimeters. (d) Explain why the geometric area of the image is given by qaqdhdq (e) Carry out the integration to find the area of the image. Figure P35.30arrow_forwardConsider the lensmirror arrangement shown in Figure P35.55. There are two final image positions to the left of the lens of focal length fL. One image position is due to light traveling from the object to the left and passing through the lens. The other image position is due to light traveling to the right from the object, reflecting from the mirror of focal length fM and then passing through the lens. For a given object position p between the lens and the mirror and measured with respect to the lens, there are two separation distances d between the lens and mirror that will cause the two images described above to be at the same location. Find both positions.arrow_forwardWhy is the following situation impossible? Consider the lensmirror combination shown in Figure P35.55. The lens has a focal length of fL = 0.200 m, and the mirror has a focal length of fM = 0.500 m. The lens and mirror are placed a distance d = 1.30 m apart, and an object is placed at p = 0.300 m from the lens. By moving a screen to various positions to the left of the lens, a student finds two different positions of the screen that produce a sharp image of the object. One of these positions corresponds to light leaving the object and traveling to the left through the lens. The other position corresponds to light traveling to the right from the object, reflecting from the mirror and then passing through the lens. Figure P35.55 Problem 55 and 57.arrow_forward
- An observer to the right of the mirror-lens combination shown in Figure P36.89 (not to scale) sees two real images that are the same size and in the same location. One image is upright, and the other is inverted. Both images are 1.50 times larger than the object. The lens has a focal length of 10.0 cm. The lens and mirror are separated by 40.0 cm. Determine the focal length of the mirror.arrow_forwardIf Joshs face is 30.0 cm in front of a concave shaving mirror creating an upright image 1.50 times as large as the object, what is the mirrors focal length? (a) 12.0 cm (b) 20.0 cm (c) 70.0 cm (d) 90.0 cm (e) none of those answersarrow_forwardA small convex mirror and a large concave mirror are separated by 1.00 m, and an object is placed 1.40 m to the left of the concave mirror (Fig. P37.69). The concave mirror forms an image of this object at distance di = 25.0 cm. This image is then reflected in the convex mirror, which forms an image a distance of 8.00 cm behind the convex mirror. What is the focal length of the small convex mirror? FIGURE P37.69arrow_forward
- Figure P23.28 shows a curved surface separating a material with index of refraction n1 from a material with index n2 . The surface forms an image I of object o. The ray shown in red passes through the surface along a radial line. Its angles of incidence and refraction are both zero, so its direction does not change at the surface. For the ray shown in blue, the direction changes according to n1 sin θ1 = n2 sin02 . For paraxial rays, we assume θ1 and θ2 are small, so we may write n1 tan θ1 = n2 tan θ2. The magnification is defined as M =h′/h. Prove that the magnification is given by M = −n1 q/n2p.arrow_forwardA dentist wishes to inspect a patient's tooth with a magnifying mirror. She places the mirror 1.00 cm behind the tooth. This results in an upright, virtual image of the tooth that is 9.00 cm behind the mirror. What is the mirror's radius of curvature (in cm)? What magnification describes the image described in this passage?arrow_forwardAn Unidentified Flying Object. You and your team spot a strange object hovering over the open ocean in broad daylight. You have nothing to reference in order to estimate its distance, height, or size, but you get the idea to use a concave mirror from a signal light. You disassemble the light, extract the mirror, and point it at the sun. You measure the distance to the sun's image from the center of the mirror to be 3.406 m. You then point the mirror at the object and find that its image forms 3.441 m from the center of the mirror. The angle between the line of sight to the object and the horizontal is 28.5 ° (a) What is the distance between you and the object? (b) What is the height of the object above the ground? (c) If the width of the object's image is 11.8 cm, what is the width of the object? (a) Number 1 (b) Number (c) Number i Units Units Unitsarrow_forward
- A spider hangs by a strand of silk at an eye level 34 cm in front of a plane mirror. The distance between you and the spider is 68 cm. Find the distance between your eye and the image of the spider. options: 68 cm 102 cm 136 cm 125 cmarrow_forwardA = 60 B = 70 C = 170 D = 7arrow_forward30. In Figure P35.30, a thin converging lens of focal length QIC 14.0 cm forms an image of the square abcd, which is h, h, = 10.0 cm high and lies between distances of p, = 20.0 cm and p. = 30.0 cm from the lens. Let a', b', c', and d' represent the respective corners of the image. Let q, rep- resent the image distance for points a' and b', q. represent the image dis- tance for points e' and d', , represent the dis- tance from point b' to the axis, and h represent the height of c'. (a) Find g. Ie h, and h'. (b) Make a sketch of the image. (c) The area of the object is 100 cm?. By carrying out the following steps, you will evaluate the area of the image. Let q represent the image distance of any point between a' and d', for which the object distance is p. Let k' represent the dis- tance from the axis to the point at the edge of the image between b' and c' at image distance q. Demonstrate that a dF a Figure P35.30 |시= 10.0g 14.0arrow_forward
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