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Ch 34: Ray Optics
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796당신이 사용하는 게 아니라요?교과서 변경
모든 교과서Knight Calc 5th EditionCh 34: Ray Optics문제 50
34장, 문제 50

A horizontal meter stick is centered at the bottom of a 3.0-m-deep, 3.0-m-wide pool of water. Suppose you place your eye just above the edge of the pool and look along the direction of the meter stick. What angle do you observe between the two ends of the meter stick if the pool is (a) empty and (b) completely filled with water?

검증된 단계별 안내
1
Determine the geometry of the problem: The meter stick is horizontal and centered at the bottom of the pool. The pool is 3.0 m wide and 3.0 m deep. Your eye is positioned just above the edge of the pool, looking along the direction of the meter stick. The angle observed between the two ends of the meter stick depends on whether the pool is empty or filled with water.
For part (a), when the pool is empty: The light rays from the two ends of the meter stick travel in straight lines to your eye. Use simple trigonometry to calculate the angle. The horizontal distance from the center of the meter stick to the edge of the pool is 1.5 m (half the width of the pool). The vertical distance from the bottom of the pool to your eye is 3.0 m. The angle θ can be calculated using the formula: θ = 2 × tan⁻¹(1.53.0).
For part (b), when the pool is filled with water: Refraction occurs at the water-air interface. Use Snell's law to account for the bending of light. The refractive index of water is approximately 1.33. The light rays from the ends of the meter stick bend as they exit the water. First, calculate the apparent depth of the meter stick using the formula: dapparent = dn, where d is the actual depth (3.0 m) and n is the refractive index of water.
Next, calculate the angle θ for the filled pool. The apparent depth reduces the vertical distance to the meter stick. Use the same trigonometric formula as in part (a), but replace the actual depth with the apparent depth: θ = 2 × tan⁻¹(1.5dapparent).
Summarize the results: The angle observed in part (a) is larger because there is no refraction. In part (b), the angle is smaller due to the bending of light caused by refraction at the water-air interface. This demonstrates how the refractive index of a medium affects the apparent position of objects underwater.

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Refraction of Light

Refraction is the bending of light as it passes from one medium to another, caused by a change in its speed. In this scenario, light travels from air into water, which has a different refractive index. This bending alters the perceived position of objects submerged in water, affecting the angle observed between the ends of the meter stick.
추천 영상:
03:46
Index of Refraction

Snell's Law

Snell's Law describes the relationship between the angles of incidence and refraction when light passes through different media. It is mathematically expressed as n1 * sin(θ1) = n2 * sin(θ2), where n is the refractive index and θ is the angle. This law is crucial for calculating the angle observed in the water-filled pool compared to when it is empty.
추천 영상:
07:59
Snell's Law

Visual Angle

The visual angle is the angle formed at the observer's eye by lines extending to the two ends of an object. It is influenced by the distance to the object and the object's size. In this problem, the visual angle changes based on whether the meter stick is in air or submerged in water, due to the effects of refraction.
추천 영상:
01:04
Critical Angle
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