/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 A person is lying on a diving bo... [FREE SOLUTION] | 91Ó°ÊÓ

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A person is lying on a diving board 3.00 m above the surface of the water in a swimming pool. The person looks at a penny that is on the bottom of the pool directly below her. The penny appears to the person to be a distance of 8.00 \(\mathrm{m}\) from her. What is the depth of the water at this point?

Short Answer

Expert verified
The depth of the water is 6.65 m.

Step by step solution

01

Understand the Refraction Concept

Light bends when it moves from water to air due to refraction. This bending causes objects underwater to appear closer than they actually are. The apparent depth and real depth can be linked using the refractive index of water.
02

Apply Snell's Law

The relationship between the apparent depth (d_a) and the real depth (d_r) is given by the formula: \(d_a = \frac{d_r}{n}\), where \(n ≈ 1.33\) is the refractive index of water. We need to use this formula to find the real depth.
03

Find Apparent Distance to the Penny

The person is 3 m above the water, and the penny appears to be 8 m away. This means the apparent depth as seen by the person is the distance from the water surface to the penny: \(d_a = 8 \text{ m} - 3 \text{ m} = 5 \text{ m}\).
04

Solve for Actual Depth

Use the equation \(d_a = \frac{d_r}{n}\) to find the real depth \(d_r\): \(5 = \frac{d_r}{1.33}\). Multiply both sides by 1.33 to get \(d_r = 5 \times 1.33 = 6.65\text{ m}\).
05

Conclusion

The calculation shows that the real depth of water at this point, taking into account the refraction, is \(6.65 \text{ m}\).

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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 describes how light bends or refracts when it moves between different mediums. Imagine light as a traveler crossing borders — each time it enters a new region, it changes direction. The law is captured in the equation \( n_1 \sin \theta_1 = n_2 \sin \theta_2 \), where \( n_1 \) and \( n_2 \) are the refractive indices of the two media, and \( \theta_1 \) and \( \theta_2 \) are the angles of incidence and refraction, respectively. This relationship allows us to determine how much the light will bend.
Applying Snell's Law helps us understand why objects under water appear closer than they really are. When light travels from water to air, it slows down, resulting in a decrease in its path distance. Hence, the penny at the pool's bottom seems nearer because of this bending effect, which is reflected through the refractive index of water.
Refractive Index
The refractive index, often represented as \( n \), is a dimensionless number that describes how fast light travels in a material compared to its speed in a vacuum. Think of it as a "light speed limit" specific to each medium. This index determines the degree to which light bends when entering a new medium.
For water, the refractive index is approximately \( 1.33 \). This means that light travels 1.33 times slower in water than in a vacuum. A higher refractive index indicates a greater bending effect of light entering at an angle.
  • If \( n > 1 \), light bends towards the normal line when entering the medium.
  • If \( n < 1 \), light bends away from the normal line.
  • If \( n = 1 \), light travels without bending.

Understanding the refractive index is crucial for predicting how deep something appears when looking through water. In the pool exercise, this principle helps calculate the apparent and real depths of the penny, helping to understand and quantify this optical illusion.
Apparent Depth vs Real Depth
When you gaze into a pool at the bottom objects, you might not see them at their true location. This discrepancy in observed position is due to refraction. The apparent depth is the perceived depth from the viewer's perspective, while the real depth is the actual depth of the object under water.
To bridge the gap between what you see and reality, we use the formula \( d_a = \frac{d_r}{n} \). Here, \( d_a \) is the apparent depth, \( d_r \) is the real depth, and \( n \) is the refractive index of the medium (in this case, water).
  • The apparent depth is always less than the real depth because light bends as it moves from water to air.
  • Knowing the apparent depth (what is seen) along with the refractive index helps precisely calculate the real depth (the actual measure).
In the given exercise, solving for the penny's real position required determining this difference. With the refractive index of water, we found that although the penny seemed only 5 meters below the surface when viewed from the diving board, its true depth was actually around 6.65 meters.

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Most popular questions from this chapter

A Glass Rod. Both ends of a glass rod with index of refraction 1.60 are ground and polished to convex hemispherical surfaces. The radius of curvature at the left end is \(6.00 \mathrm{cm},\) and the radius of curvature at the right end is 12.0 \(\mathrm{cm} .\) The length of the rod between vertices is 40.0 \(\mathrm{cm} .\) The object for the surface at the left end is an arrow that lies 23.0 \(\mathrm{cm}\) to the left of the vertex of this surface. The arrow is 1.50 \(\mathrm{mm}\) tall and at right angles to the axis. (a) What constitutes the object for the surface at the right end of the rod? (b) What is the object distance for this surface? (c) Is the object for this surface real or virtual? (d) What is the position of the final image? (e) Is the final image real or virtual? Is it erect or inverted with respect to the original object? (f) What is the height of the final image?

cp calc You are in your car driving on a highway at 25 \(\mathrm{m} / \mathrm{s}\) when you glance in the passenger-side mirror (a convex mirror with radius of curvature 150 \(\mathrm{cm}\) ) and notice a truck approaching. If the image of the truck is approaching the vertex of the mirror at a speed of 1.9 \(\mathrm{m} / \mathrm{s}\) when the truck is 2.0 \(\mathrm{m}\) from the mirror, what is the speed of the truck relative to the highway?

A glass rod with a refractive index of 1.55 is ground and polished at both ends to hemispherical surfaces with radii of 6.00 \(\mathrm{cm} .\) When an object is placed on the axis of the rod, 25.0 \(\mathrm{cm}\) to the left of the left-hand end, the final image is formed 65.0 \(\mathrm{cm}\) to the right of the right-hand end. What is the length of the rod measured between the vertices of the two hemispherical surfaces?

Recall that the intensity of light reaching film in a camera is proportional to the effective area of the lens. Camera A has a lens with an aperture diameter of 8.00 \(\mathrm{mm} .\) It photographs an object using the correct exposure time of \(\frac{1}{30} \mathrm{s}\) . What exposure time should be used with camera \(\mathrm{B}\) in photographing the same object with the same film if this camera has a lens with an aperture diameter of 23.1 \(\mathrm{mm}\) ?

A concave mirror has a radius of curvature of 34.0 \(\mathrm{cm} .\) (a) What is its focal length? (b) If the mirror is immersed in water (refractive index 1.33), what is its focal length?

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