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\(\cdot\) To a person swimming 0.80 \(\mathrm{m}\) beneath the surface of the water in a swimming pool, the diving board directly overhead appears to be a height of 5.20 \(\mathrm{m}\) above the swimmer. What is the actual height of the diving board above the surface of the water?

Short Answer

Expert verified
The actual height of the diving board above the water is approximately 3.11 meters.

Step by step solution

01

Understanding Snell's Law

The apparent depth of an object submerged in water is different from its actual depth when viewed from above due to refraction. The relationship between apparent depth and actual depth can be determined by Snell's Law, which relates the indices of refraction of two media.
02

Applying Apparent Depth Formula

Use the apparent depth formula: \( d_{apparent} = d_{actual} \times \frac{n_{water}}{n_{air}} \). Given that \( d_{apparent} = 5.20 \text{ m} \) and \( n_{water} = 1.33 \), \( n_{air} = 1.00 \), solve for \( d_{actual} \).
03

Calculating Actual Depth

Rearrange and solve \( d_{actual} = \frac{d_{apparent} \times n_{air}}{n_{water}} \). Substitute the values to get \( d_{actual} = \frac{5.20 \times 1.00}{1.33} \approx 3.91 \text{ m} \).
04

Determine Height Above Water

To find the height of the diving board above the water surface, subtract the swimmer's depth from the actual depth: \( 3.91 \text{ m} - 0.80 \text{ m} = 3.11 \text{ m} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Snell's Law
Snell's Law is a fundamental principle in optics that explains how light bends, or refracts, when it travels from one medium to another. This change of direction occurs due to the difference in the speed of light in different materials. Snell's Law is expressed by the formula:
  • \( n_1 \sin\theta_1 = n_2 \sin\theta_2 \)
Here, \( n_1 \) and \( n_2 \) are the indices of refraction of the two media, and \( \theta_1 \) and \( \theta_2 \) are the angles of incidence and refraction, respectively.
This law is crucial in understanding how light behaves when it hits a surface at an angle that is not perpendicular. In practical terms, it helps explain phenomena like why a swimming pool looks shallower than it actually is.
By using Snell's Law, we can find out how light bends when moving from air to water or vice versa, which is essential when determining apparent and actual depths.
apparent depth
Apparent depth is a concept that helps us understand how objects submerged in a transparent medium, like water, appear to be closer to the surface than they actually are. This optical illusion is due to the refraction of light rays as they move from water (a denser medium) to air (a less dense medium).
To calculate apparent depth, we use the formula:
  • \( d_{apparent} = d_{actual} \times \frac{n_{water}}{n_{air}} \)
Where \( d_{apparent} \) is what appears to be the depth to an observer, \( d_{actual} \) is the true depth of the object, and \( n_{water} \) and \( n_{air} \) are the indices of refraction for water and air, respectively.
This formula is particularly useful in scenarios like swimming pools, where understanding the true positions of objects can be essential for safety and design purposes.
indices of refraction
Indices of refraction, or refractive indices, are numbers that describe how light propagates through a particular medium. The higher the index, the slower light travels within that medium.
  • For air, the index of refraction is generally around \( 1.00 \), indicating that light travels at nearly its maximum speed.
  • For water, the index is approximately \( 1.33 \), meaning light slows down by about 33% compared to in a vacuum.
These indices are crucial for calculating how much the light will bend when transitioning between mediums, as seen in Snell's Law. Knowing the indices of refraction helps in practical applications like designing lenses, understanding underwater imaging, and even in everyday scenarios like swimming pool design to correctly estimate object positions.

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Most popular questions from this chapter

\(\cdot\) A converging lens with a focal length of 7.00 \(\mathrm{cm}\) forms an image of a 4.00 -mm-tall real object that is to the left of the lens. The image is 1.30 \(\mathrm{cm}\) tall and erect. Where are the object and image located? Is the image real or virtual?

An object is placed between two plane mirrors arranged at right angles to each other at a distance \(d_{1}\) from the surface of one mirror and a distance \(d_{2}\) from the surface of the other. (a) How many images are formed? Show the location of the images in a diagram.

If you place a concave mirror with a focal length of 1 \(\mathrm{m}\) into a liquid that has an index of refraction of \(3,\) what will happen? A. The mirror will no longer be able to focus light. B. The focal length of the mirror will decrease. C. The focal length of the mirror will increase. D. Nothing will happen.

A converging lens has a focal length of 14.0 \(\mathrm{cm} .\) For each of two objects located to the left of the lens, one at a distance of 18.0 \(\mathrm{cm}\) and the other at a distance of \(7.00 \mathrm{cm},\) determine (a) the image position, (b) the magnification, (c) whether the image is real or virtual, and (d) whether the image is erect or inverted. Draw a principal-ray diagram in each case.

\(\cdot\) The lens of the eye. The crystalline lens of the human eye is a double-convex lens made of material having an index of refraction of 1.44 (although this varies). Its focal length in air is about 8.0 \(\mathrm{mm}\) , which also varies. We shall assume that the radii of curvature of its two surfaces have the same magnitude. (a) Find the radii of curvature of this lens. (b) If an object 16 \(\mathrm{cm}\) tall is placed 30.0 \(\mathrm{cm}\) from the eye lens, where would the lens focus it and how tall would the image be? Is thisimage real or virtual? Is it erect or inverted? (Note: The results.obtained here are not strictly accurate, because the lens is embedded in fluids having refractive indexes different from that of air.)

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