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Chapter 34: Problem 17

A person swimming 0.80 \(\mathrm{m}\) below the surface of the water in a swimming pool looks at the diving board that is directly overhead and sees the image of the board that is formed by refraction at the surface of the water. This image is 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 is 6.116 m above the water surface.

Step by step solution

01

Understand the concept

This problem involves refraction, which is the bending of light as it passes from one medium to another. We are given the apparent height seen by the swimmer and need to find the actual height using Snell's Law.
02

Identify the knowns and unknowns

- Given: Apparent height of image above water surface: 5.20 m; Depth of swimmer below water surface: 0.80 m - Unknown: Actual height of the diving board above the water surface (real height).
03

Use Snell's Law for a level surface

Snell's Law relates the angle of incidence (i) and the angle of refraction (r) through the equation: \ \(n_1 \sin(i) = n_2 \sin(r)\).However, in terms of apparent depth (d') and real depth (d) for normal viewing angles from water to air, we can approximate: \ \(\frac{\text{apparent height}}{\text{real height} + \text{swimmer's depth}} = \frac{n_{air}}{n_{water}}\) \ where \(n_{air}\) is about 1.00 and \(n_{water}\) is about 1.33.
04

Set up the equation

Given the equation: \[\frac{5.20}{h + 0.80} = \frac{1.00}{1.33}\]where \(5.20\) m is the apparent height, \(h\) is the real height above the water, and the refractive indices are for air and water respectively.
05

Solve the equation for real height (h)

Rearrange the equation and solve for \(h\): \[5.20 \times 1.33 = (h + 0.80) \times 1.00\]\[6.916 = h + 0.80\]Subtract 0.80 from both sides:\[h = 6.916 - 0.80 = 6.116 \text{ m}\]
06

Interpretation and Conclusion

The calculated real height of the diving board above the surface of the water is 6.116 m. Therefore, the height that was seen by the swimmer is affected by the bending of light as it exited the water.

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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 that explains the refraction of light. When light travels from one medium to another, its speed changes, causing it to bend. Snell's Law provides a way to calculate this bending. The formula is:\[ n_1 \sin(i) = n_2 \sin(r) \]where:
  • \( n_1 \) and \( n_2 \) are the refractive indices of the two media.
  • \( i \) is the angle of incidence, and \( r \) is the angle of refraction.
This law helps us understand how objects appear displaced when viewed through different mediums, such as water and air. It is crucial in designing lenses and understanding optical phenomena, like the apparent depth perceived by a swimmer in a pool.
Refractive Index
The refractive index is a measure of how much light bends when it enters a material. It is a dimensionless number that describes how fast light travels through the medium compared to vacuum. The refractive index for a material is shown as \( n = \frac{c}{v} \), where:
  • \( c \) is the speed of light in vacuum.
  • \( v \) is the speed of light in the medium.
For instance, the refractive index of air is about 1.00, while for water it is around 1.33. This difference means light slows down and bends more when entering water from air. The refractive index is key to solving problems involving apparent and real depths, as is shown in the exercise.
Apparent Depth
Apparent depth is how deep an object appears when viewed from above the surface of a different medium. It results from refraction, where light rays bend upwards as they move from a denser medium like water to a less dense medium like air. To calculate apparent depth, the formula used is:\[ \frac{\text{apparent depth}}{\text{real depth}} = \frac{n_{air}}{n_{water}} \]In our exercise, the apparent depth helps determine how much the image of the diving board is shifted by refraction. This visible shift creates the illusion that objects under water seem closer to the surface than they actually are.
Real Depth
Real depth refers to the true distance of an object beneath the surface of a medium. Unlike apparent depth, real depth does not consider the effects of light bending. In solving our exercise, finding the real height or depth involves using apparent depth, refractive indices, and the swimmer's position below the water surface.By rearranging the refractive relationship, we were able to solve:\[ h = 6.916 - 0.80 = 6.116 \text{ m} \]This calculation provides the true height of the diving board above the water. Understanding the difference between apparent and real depths is crucial for underwater navigation and imaging.

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Most popular questions from this chapter

You want to view an insect 2.00 \(\mathrm{mm}\) in length through a magnifier. If the insect is to be at the focal point of the magnifier, what focal length will give the image of the insect an angular size of 0.025 radian?

The Cornea As a Simple Lens. The cornea behaves as a thin lens of focal length approximately \(1.8 \mathrm{cm},\) although this varies a bit. The material of which it is made has an index of refraction of \(1.38,\) and its front surface is convex, with a radius of curvature of 5.0 \(\mathrm{mm} .\) (a) If this focal length is in air, what is the radius of curvature of the back side of the cornea? (b)The closest distance at which a typical person can focus on an object (called the near point) is about \(25 \mathrm{cm},\) although this varies considerably with age. Where would the cornea focus the image of an 8.0 -mm- tall object at the near point? (c) What is the height of the image in part (b)? Is this image real or virtual? Is it erect or inverted? (Note: The results obtained here are not strictly accurate because, on one side, the cornea has a fluid with a refractive index different from that of air.)

A camera with a 90 -mm-focal-length lens is focused on an object 1.30 \(\mathrm{m}\) from the lens. To refocus on an object 6.50 \(\mathrm{m}\) from the lens, by how much must the distance between the lens and the film be changed? To refocus on the more distant object, is the lens moved toward or away from the film?

A layer of benzene \((n=1.50) 4.20 \mathrm{cm}\) deep floats on water \((n=1.33)\) that is 6.50 \(\mathrm{cm}\) deep. What is the apparent distance from the upper benzene surface to the bottom of the water layer when it is viewed at normal incidence?

A converging lens with a focal length of 12.0 \(\mathrm{cm}\) forms a virtual image 8.00 \(\mathrm{mm}\) tall, 17.0 \(\mathrm{cm}\) to the right of the lens. Determine the position and size of the object. Is the image erect or inverted? Are the object and image on the same side or opposite sides of the lens? Draw a principal-ray diagram for this situation.
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