Concept explainers
Figure 34-25 shows a fish and a fish stalker in water. (a) Does the stalker see the fish in the general region of point a or point b? (b) Does the fish see the (wild) eyes of the stalker in the general region of point c or point d?
Figure 34-25 Question 1.
To find:
a) Whether the stalker sees the fish in the general region of point a or point b.
b) Whether the fish sees the eyes of the stalker in the general region of point c or point d.
Answer to Problem 1Q
Solution:
a) The stalker sees the fish in the general region of point a.
b) The fish sees the eyes of the stalker in the general region of point c.
Explanation of Solution
1) Concept:
Analyzing the given figure and using concept of refraction we can find the region in which the stalker sees the fish and the fish sees the eyes of the stalker.
2) Given:
Fig. 34-25.
3) Calculations:
a) The ray diagram:
When light travels from rarer to denser medium, it refracts towards the normal and when it travels from denser to rarer medium, it refracts away from the normal.
From the above ray diagram, we can conclude that, the ray of light is travelling from water to air, that is, from the denser medium to the rarer medium. Hence, the light should refract away from the normal. That is,
From the figure, we can conclude that this is possible only in region a.
Therefore, the stalker sees the fish in the general region of point a.
b) The ray diagram:
From the above ray diagram, we can conclude that the ray of light is travelling from air to water, that is, from the rarer medium to the denser medium. Hence, the light should refract towards the normal. That is,
From the figure, we can conclude that, this is possible only in region c.
Therefore, the fish see the eyes of the stalker in the general region of point c.
Conclusion:
When the light travels from rarer to denser medium, it refracts towards the normal and when it travels from denser to rarer medium, it refracts away from the normal.
Want to see more full solutions like this?
Chapter 34 Solutions
Fundamentals of Physics Extended
Additional Science Textbook Solutions
Campbell Biology in Focus (2nd Edition)
Campbell Essential Biology with Physiology (5th Edition)
Chemistry
Microbiology: An Introduction
Chemistry: An Introduction to General, Organic, and Biological Chemistry (13th Edition)
Biochemistry: Concepts and Connections (2nd Edition)
- Suppose you are looking down at a highway from a jetliner flying at an altitude of 6.0 km. How far apart must two cars be if you are able to distinguish them? Assume that =550 nm and that the diameter of your pupils is 4.0 mm.arrow_forwardFigure 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 1 = n2 sin 2. For paraxial rays, we assume 1 and 2 are small, so we may write n1 tan 1 n2 tan 2. The magnification is defined as M = h/h. Prove that the magnification is given by M = n1q/n2p. Figure P23.28arrow_forward1 Figure 34-25 shows a fish and a fish stalker in water. (a) Does the stalker see the fish in the general re- gion of point a or point b? (b) Does the fish see the (wild) eyes of the stalker in the general region of point cor point d?arrow_forward
- In the figure, light is incident at angle 6, = 41.0° on a boundary between two transparent materials. Some of the light travels down through the next three layers of transparent materials, while some of it reflects upward and then escapes into the air. If n1 = 1.28, n, = 1.38, ng = 1.30 and n4 = 1.43, what is the value of (a) 05 and (b) 04? Air (a) 05 = NumberT32.02 UnitsTo (degrees) (b)04 = NumberT43.1 Units (degrees)arrow_forwardRed light is incident in air on a 30° -60° - 90 ° prism as shown. The incident beam is directed at an angle of ₁ = 40.20 with respect to the horizontal and enters the prism at a height h = 20 cm above the base. The beam leaves the prism to the air at a distance d = 54.2 along the base as shown. 45.66 1.36 30⁰ 1) What is 2, the angle the beam in the prism makes with the horizontal axis? 71.9 d degrees Submit 2) What is n, the index of refraction of the prism for red light? dviolet d Submit Submit 02 3) What is 3, the angle the transmitted beam makes with the horizontal axis? h degrees Submit 03 degrees Submit (+) 4) What is 41,max, the maximum value of 1 for which the incident beam experiences total internal reflection at the horizontal face of the prism? degrees Submit + + 5) The red beam is now replaced by a violet beam that is incident at the same angle 1 and same height h. The prism has an index of refraction nviolet = 1.46 for violet light. Compare dviolet, the exit distance for…arrow_forwardWhat is the smallest thing we can see? O smallest object that we can process with our eyes is limited to the size of the photoreceptor cells in the retina. In order for us to to distinguish any detail in an object, its image cannot be smaller cannot be smaller than a single retinal cell. Although the size depends on the type of cell (cone or rod), a diameter of of a few microns (mm) is common near the center of the eye. We should model the eye as a sphere 2.50 cm in diameter with a single slender lens in front and the retina behind, with photoreceptor cells 5.0 mm in diameter. (a) What is the smallest object object that you can perceive at a point near 25 cm? What angle is subtended by this object in the eye? Express your answer in units of minutes (1° 60 min) and compare it with the typical experimental value of about 1.0 min. (Note: there are other limitations, but we will ignore them here).arrow_forward
- Under dark conditions, the maximum diameter of a human pupil is 7.0 mm, where an owl's pupil may be 8.5 mm. Assume a human can optically resolve two closely spaced objects at a distance r. (a) By what factor could the distance between the two objects be reduced and still have the owl optically resolve them at the same distance r? (b) If the distance between the two objects remains fixed, by what factor could r be increased and still have the owl optically resolve the two objects? In both (a) and (b), assume the wavelength of the light remains constant. (a) Number i Units (b) Number i Unitsarrow_forward(a) A small light fixture on the bottom of a swimming pool is 0.92 m below the surface. The light emerging from the still water forms a circle on the water surface. What is the diameter of this circle? (Give your answer, in m, to at least two decimal places.) Xm (b) What If? If a 1.63 cm thick layer of oil (noil = 1.35) is spread uniformly over the surface of the water, what is the diameter of the circle of light emerging from the swimming pool? (Give your answer, in m, to at least two decimal places.) X marrow_forwardH 2.0m d A scuba diver training in a pool looks at his instructor as shown in the figure. The angle between the ray in the water and the perpendicular to the water is 0: 25 °arrow_forward
- Building contractors often install double-glazed windows to prevent thermal energy (heat) from entering or exiting a building. In addition to being effective insulators, such windows present interesting optical effects. glass In the figure, a double-glazed window consists of two identical panes of glass (ng = 1.44), each y, = 62.0 mm air thick, separated by an air gap of ya = 49.6 mm. If light incident on the glass makes an angle of p = 40.00° with respect to the glass, find the shift in path Ax as the light enters glass the room. Use n, = 1.00 for the index of refraction of air. Ax = mmarrow_forwardAn astronaut in a space shuttle claims she can just barely resolve two point sources on Earth’s surface, 160 km below. Calculate their (a) angular and (b) linear separation, assuming ideal conditions.Take l = 540 nm and the pupil diameter of the astronaut’s eye to be 5.0 mm.arrow_forward(V) fiber at an angle 0 with respect to the fiber axis. Show that there is a range of 0 (called "acceptance angle") outside which the light will no longer be able to propagate inside the fiber. Find the expression of the acceptance angle in terms of n1 and n2. Calculate the value for the case of ni Consider a step index optical fiber. Light enters from air at one end of the 1.5 and n2 1.49. n2 Cladding n1 Corearrow_forward
- Physics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage LearningCollege PhysicsPhysicsISBN:9781305952300Author:Raymond A. Serway, Chris VuillePublisher:Cengage LearningUniversity Physics Volume 3PhysicsISBN:9781938168185Author:William Moebs, Jeff SannyPublisher:OpenStax
- An Introduction to Physical SciencePhysicsISBN:9781305079137Author:James Shipman, Jerry D. Wilson, Charles A. Higgins, Omar TorresPublisher:Cengage Learning