Problem 28
Question
\(\cdot\) 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?
Step-by-Step Solution
Verified Answer
The actual height of the diving board above the water is approximately 3.11 meters.
1Step 1: 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.
2Step 2: 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 \( d_{actual} \).
3Step 3: Calculating Actual Depth
Rearrange and solve \( d_{actual} = \frac{d_{apparent} \times n_{air}}{n_{water}} \). Substitute the values to get \( d_{actual} = \frac{5.20 \times 1.00}{1.33} \approx 3.91 \text{ m} \).
4Step 4: 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: \( 3.91 \text{ m} - 0.80 \text{ m} = 3.11 \text{ m} \).
Key Concepts
Snell's Lawapparent depthindices of refraction
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:
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.
- \( n_1 \sin\theta_1 = n_2 \sin\theta_2 \)
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:
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.
To calculate apparent depth, we use the formula:
- \( d_{apparent} = d_{actual} \times \frac{n_{water}}{n_{air}} \)
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.
Other exercises in this chapter
Problem 26
\(\cdot\) A skin diver is 2.0 \(\mathrm{m}\) below the surface of a lake. A bird flies overhead 7.0 \(\mathrm{m}\) above the surface of the lake. When the bird
View solution Problem 27
\(\bullet \mathrm{A}\) zoo aquarium has transparent walls, so that spectators on both sides of it can watch the fish. The aquarium is 5.50 \(\mathrm{m}\) across
View solution Problem 29
\(\cdot\) 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
View solution Problem 30
A converging lens with a focal length of 90.0 \(\mathrm{cm}\) forms an image of a 3.20 -cm-tall real object that is to the left of the lens. The image is 4.50 \
View solution