COLLEGE PHYSICS
2nd Edition
ISBN: 9781464196393
Author: Freedman
Publisher: MAC HIGHER
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Chapter 23, Problem 35QAP
To determine
The ratio for light required for light to travel through 1000 m of air to the time required for light to travel through 1000 m of vacuum
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(no solution needed)A light ray from air is incident to the surface of water ( n=1.33) at an angle of 30°. Relative to the angle of incidence, the angle of reflection _________.
is smaller
is the same
is larger
may be any of the above
The angle of refraction is equal to _________.
32°
40°
22°
25°
What is the critical angle of incidence for water-air interface?
42°
36°
50°
49°
If the light ray from air is incident on the surface of a glass ( n= 1.73), what is the speed of light in the glass?
1.25 x 108 m/s
1.73 x 108 m/s
1.8 x 108 m/s
2.7 x 108 m/s
The angle of refraction in the glass is _____ degrees.
20
36
30
17
Jurassic Park - some Physics. Amber (n = 1.5343) is a transparent brown-yellow fossil resin. An insect, trapped and preserved within the amber, appears to be 19.65 mm beneath the surface when viewed directly from above. How far below the surface is the insect actually located?
A friend lends you the eyepiece of his microscope to use on your own microscope. He claims that since his eyepiece has the same diameter as yours but twice the focal length, the resolving power of your microscope will be doubled. Is his claim valid? Explain.
Chapter 23 Solutions
COLLEGE PHYSICS
Ch. 23 - Prob. 1QAPCh. 23 - Prob. 2QAPCh. 23 - Prob. 3QAPCh. 23 - Prob. 4QAPCh. 23 - Prob. 5QAPCh. 23 - Prob. 6QAPCh. 23 - Prob. 7QAPCh. 23 - Prob. 8QAPCh. 23 - Prob. 9QAPCh. 23 - Prob. 10QAP
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- A ray of monochromatic light travelling in air strikes the end of a 21 cm wide block of Plexiglas at an incident angle of65°. The index of refraction for the Plexiglas is 1.51 and for air is 1.003 The light travels to the opposite side of the block and refracts out into the air. The path of the refracted light ray isparallel to the original light ray but displaced a distance d. Calculate the value of d. No you cannot just look up aformula on internet, draw diagram.arrow_forwardSome transparent substance has refractive index 1.6. Find the critical angle for the border the substance-air. Calculate to one decimal.arrow_forwardConsider sunlight entering the Earth's atmosphere at sunrise and sunset-that is, at a 90° incident angle. Taking the boundary between nearly emptyspace and the atmosphere to be sudden, calculate the angle of refractionfor sunlight. This lengthens the time the Sun appears to be above thehorizon, both at sunrise and sunset. Now construct a problem in which youdetermine the angle of refraction for different models of the atmosphere,such as various layers of varying density. Your instructor may wish to guideyou on the level of complexity to consider and on how the index ofrefraction varies with air density.arrow_forward
- Sunlight or starlight passing through the earth’s atmosphere is always bent toward the vertical. Why? Does this mean that a star is not really where it appears to be? Explain.arrow_forwardLight travels from air toward water. 1) If the angle that is formed by the light beam in air with respect to the normal line between the two media is 25°, calculate the the angle of refraction of the light in the water. The indices of refraction of air and water are nair = 1.00029 and nwater = 1.33, respectively. (Express to 2 sig fig).arrow_forwardLight is an incident on a vertical interface between two mediums as shown in the picture. The medium on the left has an index of refraction n(left) & the medium on the right has an index of refraction n(right). The light comes from the bottom left. If n(left)=1.3, n(right)=2.3, & angle of incidence = 41degrees. What is the angle of refraction? -68.2 -49.0 -0 -41.0 -21.8 If n(left)=1.3, n(right)=2.3, & angle of incidence= 41degrees, what is the angle of reflection? *same choices as first question*arrow_forward
- 14: A fisherman spots a fish underneath the water. It appears that the fish is d0 = 0.57 m under the water surface at an angle of θa = 64 degrees with respect to the normal to the surface of the water. The index of refraction of water is nw = 1.3 and the index of refraction of air is na = 1. (a) The perpendicular distance from the apparent position of the fish to the normal of the water surface shown in the figure is L. Express L in terms of tan θa and d0. (b) Solve for the numerical value of L, in meters. (c) Express the sine of the angle θw, in terms of θa, nw, and na. (d) Solve for the numerical value of θw in degrees. (e) Now assume that the real position of the fish is directly below the apparent position, as shown in the figure. Express the real depth of the fish, d, in terms of L and tanθw. (f) Solve for the numerical value of d, in meters.arrow_forwardA laboratory (astronomical) telescope is used to view a scale that is 300 cm from the objective, which has a focal length of 20.0 cm; the eyepiece has a focal length of 2.00 cm. Calculate the angular magnification when the telescope is adjusted for minimum eyestrain. Note: The object is not at infinity, so the simple expression m = fo/fe is not sufficiently accurate for this problem. Also, assume small angles, so that tan = .arrow_forwardA laboratory (astronomical) telescope is used to view a scale that is 300 cm from the objective, which has a focal length of 20.0 cm; the eyepiece has a focal length of 2.00 cm. Calculate the angular magnification when the telescope is adjusted for minimum eyestrain. Note: The object is not at infinity, so the simple expression m = fo/fe is not sufficiently accurate for this problem. Also, assume small angles, so that tan = .arrow_forward
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