Explanation Given data: Angle of incident of laser beam on the mirror is 30 ° . Distance between the wall and the mirror is 2 m. The mirror is been rotated by 10 ° . Explanation: Modify FIGURE P18.44 as shown in Figure 1. As, mirror is rotated by 10 ° , find the new angle of incident as follows. The new angle of incident = 30 ° + 10 ° = 40 ° According to Snell’s law, New angle of incident = New angle of reflection Substitute 40 ° for New angle of incident , New angle of reflection = 40 ° But, 40 ° is the new angle of incident for the rotated normal, which is been formed by the rotation of mirror. New angle of incident from original normal = 40 ° + 10 ° = 50 ° Write the expression of tan θ for a right angled triangle by using trigonometric rule. tan 50 ° = O P ′ d (1) And, tan 30 ° = O P ′ d (2) Substitute 2 m for d in equation (1), tan 50 ° = <

The distance laser spot move on the wall when the mirror is rotated by 10 ° .

College Physics

3rd Edition

ISBN: 9780134201962

Author: Randall D. Knight, Brian Jones, Stuart Field

Publisher: Addison-Wesley

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Question

Chapter 18, Problem 44GP

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The distance laser spot move on the wall when the mirror is rotated by 10° .

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Chapter 18, Problem 43GP

Chapter 18, Problem 45GP

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Why 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.
The object in Figure P23.52 is mid-way between the lens and the mirror, which are separated by a distance d = 25.0 cm. The magnitude of the mirrors radius of curvature is 20.0 cm, and the lens has a focal length of 16.7 cm. (a) Considering only the light that leaves the object and travels first toward the mirror, locate the final image formed by this system. (b) Is the image real or virtual? (c) Is it upright or inverted? (d) What is the overall magnification of the image? Figure P23.52
Figure P36.95 shows a thin converging lens for which the radii of curvature of its surfaces have magnitudes of 9.00 cm and 11.0 cm. The lens is in front of a concave spherical mirror with the radius of curvature R = 8.00 cm. Assume the focal points F1 and F2 of the lens are 5.00 cm from the center of the lens, (a) Determine the index of refraction of the lens material. The lens and mirror are 20.0 cm apart, and an object is placed 8.00 cm to the left of the lens. Determine (b) the position of the filial image and (c) its magnification as seen by the eye in the figure. (d) Is the final image inverted or upright? Explain.
If 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 answers
The object in Figure P23.52 is mid-way between the lens and the mirror, which are separated by a distance d = 25.0 cm. The magnitude of the mirrors radius of curvature is 20.0 cm, and the lens has a focal length of 16.7 cm. (a) Considering only the light that leaves the object and travels first toward the mirror, locate the final image formed by this system. (b) Is the image real or virtual? (c) Is it upright or inverted? (d) What is the overall magnification of the image? Figure P23.52
A 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 P37.69
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.
(i) An object is plated at a position p f from a concave mirror as shown in Figure CQ39.12a, where f is the focal length of the mirror. In a finite time interval, the object is moved to the right to a position at the focal point F of the mirror. Show that the image of the object moves at a speed greater than the speed of light. (ii) A laser pointer is suspended in a horizontal plane and set into rapid rotation as shown in Figure CQ39 12b. Show that the spot of light it produces on a distant screen can move across the screen at a speed greater than the speed of light. (If you carry out this experiment. make sure the direct laser light cannot enter a person's eyes.) (iii) Argue that the experiments in parts (i) and (ii) do not invalidate the principle that no material, no energy, and no information can move faster than light moves in a vacuum. Figure CQ39.12
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 P35.30
Curved glassair interfaces like those observed in an empty shot glass make it possible for total internal reflection to occur at the shot glasss internal surface. Consider a glass cylinder (n = 1.54) with an outer radius of 2.50 cm and an inner radius of 2.00 cm as shown in Figure P38.105. Find the minimum angle i such that there is total internal reflection at the inner surface of the shot glass. FIGURE P38.105 Problems 105 and 106.
Figure P26.72 shows a thin converging lens for which the radii of curvature of its surfaces have magnitudes of 9.00 cm and 11.0 cm. The lens is in front of a concave spherical mirror with the radius of curvature R = 8.00 cm. Assume the focal points F1 and F2 of the lens are 5.00 cm from the center of the lens. (a) Determine the index of refraction of the lens material. The lens and mirror are 20.0 cm apart, and an object is placed 8.00 cm to the left of the lens. Determine (b) the position of the final image and (c) its magnification as seen by the eye in the figure. (d) Is the final image inverted or upright? Explain.
An object represented by a gray arrow, is placed in front of a plane mirror. Which of the diagram in Figure CQ23.15 best describes the image, represented by the pink arrow? Figure CQ23.15

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