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Ch. 32 - Light: Reflection and Refraction
Giancoli Douglas - Physics for Scientists and Engineers 5th edition
Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179당신이 사용하는 게 아니라요?교과서 변경
모든 교과서Giancoli Douglas 5th editionCh. 32 - Light: Reflection and Refraction문제 93
31장, 문제 93

The paint used on highway signs often contains small transparent spheres which provide nighttime illumination of the sign’s lettering by retro-reflecting vehicle headlight beams. Consider a light ray from air incident on one such sphere of radius r and index of refraction n. Let θ be its incident angle, and let the ray follow the path shown in Fig. 32–70, so that the ray exits the sphere in the direction exactly antiparallel to its incoming direction. Considering only rays for which sin θ can be approximated as θ, determine the required value for n.

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Start by understanding the problem: The light ray enters a transparent sphere, refracts, reflects internally, and exits in a direction exactly antiparallel to its incoming direction. This is a retro-reflection phenomenon. We are tasked with finding the required value of the sphere's index of refraction (n) under the small-angle approximation (sin θ ≈ θ).
Apply Snell's Law at the point of entry of the light ray into the sphere. Snell's Law is given by: n1⁢sin⁢θ=n2⁢sin⁢θr, where n₁ is the refractive index of air (approximately 1), θ is the angle of incidence, n₂ is the refractive index of the sphere (n), and θₐ is the angle of refraction. Using the small-angle approximation, this simplifies to: θ=n⁢θr.
Next, consider the internal reflection within the sphere. The light ray travels to the opposite side of the sphere, reflects, and returns. The geometry of the sphere ensures that the total internal angle traversed by the ray is 2θₐ. For the ray to exit antiparallel to its incoming direction, the total phase shift due to refraction and reflection must align with this condition.
At the exit point, apply Snell's Law again. The ray exits the sphere back into air, so the relationship is similar to the entry point: n⁢θr=θ. This ensures that the ray exits at the same angle as it entered, but in the opposite direction.
Combine the relationships derived from Snell's Law and the geometry of the sphere. Using the small-angle approximation and the condition for retro-reflection, solve for the refractive index n. The result will depend on the geometry of the sphere and the angles involved, leading to a specific value for n that satisfies the antiparallel condition.

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주요 개념

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Refraction

Refraction is the bending of light as it passes from one medium to another with a different index of refraction. This phenomenon occurs due to the change in speed of light in different materials. The relationship between the angles of incidence and refraction is described by Snell's Law, which states that n1 * sin(θ1) = n2 * sin(θ2), where n is the index of refraction and θ is the angle of incidence or refraction.
추천 영상:
가이드 코스
03:46
Index of Refraction

Retro-reflection

Retro-reflection is the property of a surface that reflects light back to its source, regardless of the angle of incidence. This is achieved through specific geometries, such as the use of transparent spheres, which redirect incoming light rays to exit in the opposite direction. This principle is crucial for highway signs, as it enhances visibility at night by ensuring that vehicle headlights illuminate the sign effectively.
추천 영상:
가이드 코스
06:25
Law of Reflection

Index of Refraction

The index of refraction (n) is a dimensionless number that describes how much light slows down in a medium compared to its speed in a vacuum. It is defined as n = c/v, where c is the speed of light in a vacuum and v is the speed of light in the medium. A higher index indicates that light travels slower in that medium, which affects how light rays bend when entering or exiting the material, influencing the design of retro-reflective materials.
추천 영상:
가이드 코스
03:46
Index of Refraction
관련 실천
교과서 질문

Suppose Fig. 32–37 shows a cylindrical rod whose end has a radius of curvature R = 2.0 cm, and the rod is immersed in water with index of refraction of 1.33. The rod has index of refraction 1.49. Find the location and height of the image of an object 2.0 mm high located 23 cm away from the rod.

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교과서 질문

A coin lies at the bottom of a 0.95-m-deep pool. If a viewer sees it at a 45° angle, where is the image of the coin, relative to the coin? [Hint: The image is found by tracing back to the intersection of two rays.]

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교과서 질문

You hold a small flat mirror 0.50 m in front of you and can see your reflection twice in that mirror because there is a full-length mirror 1.0 m behind you (Fig. 32–71). Determine the distance of each image from you.

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교과서 질문

Figure 33–51 was taken from the NIST Laboratory (National Institute of Standards and Technology) in Boulder, CO, 2.0 km from the hiker in the photo. The Sun’s image was 15 mm across on the film. Estimate the focal length of the camera lens (actually a telescope). The Sun has diameter 1.4 x 106 km, and it is 1.5 x 108 km away.


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