Ch 34: Geometric Optics

Young & Freedman Calc - University Physics 15th Edition

Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Not the one you use?Change textbook

Young & Freedman Calc 15th Edition

Ch 34: Geometric Optics

Problem 22

Chapter 33, Problem 22

A person is lying on a diving board 3.00 m above the surface of the water in a swimming pool. She looks at a penny that is on the bottom of the pool directly below her. To her, the penny appears to be a distance of 7.00 m from her. What is the depth of the water at this point?

Verified step by step guidance

Understand the problem: The apparent depth of the penny is given as 7.00 m, and the person is 3.00 m above the water surface. The goal is to find the actual depth of the water using the concept of refraction and the refractive index of water.

Recall the formula for apparent depth: The relationship between the apparent depth \(d_a\) and the real depth \(d_r\) is given by \(d_a = \frac{d_r}{n}\), where \(n\) is the refractive index of water (approximately 1.33 for visible light).

Rearrange the formula to solve for the real depth \(d_r\): \(d_r = d_a \cdot n\). Substitute \(d_a = 7.00 \ \text{m}\) and \(n = 1.33\) into the equation.

Account for the height of the person above the water: The total depth of the water is the real depth \(d_r\) minus the height of the person above the water surface (3.00 m).

Combine the results: Add the height of the person above the water to the calculated real depth \(d_r\) to determine the total depth of the water at the point where the penny is located.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Refraction of Light

Refraction is the bending of light as it passes from one medium to another, such as from air to water. This bending occurs because light travels at different speeds in different materials. In this scenario, the apparent depth of the penny is affected by the refraction of light, making it appear deeper than its actual position.

Apparent Depth

Apparent depth refers to the perceived distance of an object submerged in a fluid, which differs from its actual depth due to refraction. When viewing an object underwater from above, the light rays bend, causing the object to appear closer to the surface than it truly is. This concept is crucial for understanding the discrepancy between the actual depth of the penny and how deep it appears to the observer.

Snell's Law

Snell's Law describes the relationship between the angles of incidence and refraction when light passes between two different media. It is mathematically expressed as n1 * sin(θ1) = n2 * sin(θ2), where n represents the refractive indices of the media. This law helps calculate the apparent depth and is essential for solving the problem of determining the actual depth of the water based on the observed distance.

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Snell's Law

Related Practice

Textbook Question

Dental Mirror. A dentist uses a curved mirror to view teeth on the upper side of the mouth. Suppose she wants an erect with a magnification of 2.00 when the mirror is 1.25 cm from a tooth. (Treat this problem as though the object and lie along a straight line.) What must be the focal length and radius of curvature of this mirror?

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Textbook Question

The glass rod of Exercise 34.22 is immersed in oil (n = 1.45). An object placed to the left of the rod on the rod's axis is to be d 1.20 m inside the rod. How far from the left end of the rod must the object be located to form the image?

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Textbook Question

The left end of a long glass rod 8.00 cm in diameter, with an index of refraction of 1.60, is ground and polished to a convex hemispherical surface with a radius of 4.00 cm. An object in the form of an arrow 1.50 mm tall, at right angles to the axis of the rod, is located on the axis 24.0 cm to the left of the vertex of the convex surface. Find the position and height of the of the arrow formed by paraxial rays incident on the convex surface. Is the erect or inverted?

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Textbook Question

The thin glass shell shown in Fig. E34.15 has a spherical shape with a radius of curvature of 12.0 cm, and both of its surfaces can act as mirrors. A seed 3.30 mm high is placed 15.0 cm from the center of the mirror along the optic axis, as shown in the figure. Calculate the location and height of the of this seed.

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Textbook Question

A Spherical Fish Bowl. A small tropical fish is at the center of a water-filled, spherical fish bowl 28.0 cm in diameter. Find the apparent position and magnification of the fish to an observer outside the bowl. The effect of the thin walls of the bowl may be ignored.

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Textbook Question

A spherical, concave shaving mirror has a radius of curvature of 32.0 cm. Where is the image? Is the image real or virtual?

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