(a)
The speed of the laser beam when the spot of light on the wall is at distance
(a)
Answer to Problem 59P
The speed of the laser beam when the spot of light on the wall is at distance
Explanation of Solution
Let
From the Figure, write the expression for
Here,
Solve equation (I) for
Differentiate the equation (II) with respect to time, to find the speed of the laser spot at the wall.
From the Figure, using Pythagorean write the expression for
`
If the mirror turns through angle
Use equation (IV) and (V) in (III).
Conclusion:
Therefore, the speed of the laser beam when the spot of light on the wall is at distance
(b)
The value of the
(b)
Answer to Problem 59P
The value of the
Explanation of Solution
The speed of the laser beam when the spot of light on the wall is at distance
Conclusion:
Therefore, the value of the
(c)
The minimum value for the speed.
(c)
Answer to Problem 59P
The minimum value for the speed is
Explanation of Solution
The speed of the laser beam when the spot of light on the wall is at distance
Substitute,
Conclusion:
Therefore, the minimum value for the speed is
(d)
The maximum speed of the spot on the wall.
(d)
Answer to Problem 59P
The maximum speed of the spot on the wall is
Explanation of Solution
The maximum speed occurs when the reflected beam arrives at the corner of the room.
Substitute,
Conclusion:
Therefore, the maximum speed of the spot on the wall is
(e)
The time interval in which the spot changes from its minimum to maximum speed.
(e)
Answer to Problem 59P
The time interval in which the spot changes from its minimum to maximum speed is
Explanation of Solution
The reflected laser beam rotates through
The time interval between the minimum and maximum speed can b e calculated as follows,
Conclusion:
Substitute,
Therefore, the time interval in which the spot changes from its minimum to maximum speed is
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Chapter 25 Solutions
Principles of Physics
- 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 P23.52arrow_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_forwardFigure 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.arrow_forward
- 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 P23.52arrow_forward(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 CQ39.12arrow_forwardIn 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_forward
- Two converging lenses having focal lengths of f1 = 10.0 cm and f2 = 20.0 cm are placed a distance d = 50.0 cm apart as shown in Figure P35.48. The image due to light passing through both lenses is to be located between the lenses at the position x = 31.0 cm indicated. (a) At what value of p should the object be positioned to the left of the first lens? (b) What is the magnification of the final image? (c) Is the final image upright or inverted? (d) Is the final image real or virtual?arrow_forwardAn 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_forwardConsider a light ray that enters a pane of glass with air on one side and water on the other side as shown in Figure P38.21. The light ray experiences refraction at the first interface when it enters the glass from the water and again at the second interface when it exits the glass into the air. Assume the index of refraction of the glass is 1.54. For a ray of light, find the angle of incidence 1 in the water such that the ray experiences total internal reflection when it strikes the glassair interface on the other side. FIGURE P38.21arrow_forward
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