/*! 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 15 A speck of dirt is embedded 3.50... [FREE SOLUTION] | 91Ó°ÊÓ

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A speck of dirt is embedded 3.50 \(\mathrm{cm}\) below the surface of a sheet of ice \((n=1.309) .\) What is its apparent depth when viewed at normal incidence?

Short Answer

Expert verified
The apparent depth is approximately 2.67 cm.

Step by step solution

01

Identify the Known Quantities

We know that the actual depth of the speck of dirt is 3.50 cm below the surface, and the refractive index of ice is given as 1.309. These are the key pieces of information we need to solve the problem.
02

Apply the Apparent Depth Formula

Use the formula for apparent depth when viewed at normal incidence:\[apparent\ depth = \frac{real\ depth}{n}\]where \(n\) is the refractive index of the medium. In this instance, the real depth is 3.50 cm and \(n\) is 1.309.
03

Substitute the Known Values

Substitute the given values into the apparent depth formula:\[apparent\ depth = \frac{3.50\ \text{cm}}{1.309}\]
04

Calculate the Apparent Depth

Perform the division to find the apparent depth:\[apparent\ depth = \frac{3.50}{1.309} \approx 2.67\ \text{cm}\]Thus, the apparent depth of the speck of dirt is approximately 2.67 cm.

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Key Concepts

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

Refractive Index
When light travels from one medium to another, such as from air into ice, its speed changes. This change in speed causes the light to bend or refract. The refractive index is a number that describes how much light bends when entering a particular medium. It is denoted by the letter \(n\), which stands for "index of refraction".
The refractive index has a very practical application. It helps us determine how much light slows down compared to its speed in a vacuum. For example, a refractive index \(n=1.309\) means light travels slower in ice than in air, as air has a refractive index of about 1. Key concepts about refractive index include:
  • A higher refractive index indicates a denser medium and greater bending of light.
  • The refractive index of air is approximately 1, while that of vacuum is exactly 1.
  • Refractive index values vary for different materials, which helps define optical properties.
Understanding the refractive index is crucial for solving many optics problems, such as calculating apparent depth.
Normal Incidence
Normal incidence refers to a situation where a ray of light hits a surface directly from above, forming a 90-degree angle with the surface. In practical terms, it means that the light is traveling perpendicular to the surface it encounters. Why does this matter? Well, at normal incidence, the light's path is straightforward. There's no sideways bending of the light ray as it passes into the new medium. This simplifies calculations and allows us to use the apparent depth formula without additional complications. Normal incidence is significant because:
  • It ensures that light is neither bent nor deflected sideways when entering a medium.
  • The refraction effects only change the perceived depth, not the lateral position of objects.
  • Many formulas in optics, such as for apparent depth, are simplified under normal incidence conditions thanks to this lack of lateral bending.
By understanding normal incidence, students can better grasp how refraction affects the appearance of objects viewed through transparent materials like water or ice.
Optics Problem Solving
Solving optics problems involves understanding how light behaves under various conditions such as refraction, reflection, or transmission. In our example, the exercise is to say what happens to the apparent depth of an object beneath ice. To approach optics problem solving, a few strategies can help:
  • Clearly identify the known quantities, such as actual depth and refractive index.
  • Choose the correct formula. In this case, it’s the apparent depth formula \(\text{apparent depth} = \frac{\text{real depth}}{n}\).
  • Substitute the known values into the formula, making sure the units are consistent.
  • Perform the calculations step-by-step, ensuring precision and accuracy.
By breaking down the problem, using specific known values, and then calculating step-by-step, you can find reliable solutions and better understand how optics principles apply to realistic scenarios. This approach is valuable for any physics or engineering problem involving light and vision.

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Most popular questions from this chapter

The diameter of Mars is \(6794 \mathrm{km},\) and its minimum distance from the earth is \(5.58 \times 10^{7} \mathrm{km} .\) When Mars is at this distance, find the diameter of the image of Mars formed by a spherical, concave, telescope mirror with a focal length of 1.75 \(\mathrm{m}\) .

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 1.5 \(\mathrm{m} / \mathrm{s}\) when the truck is 2.0 \(\mathrm{m}\) away, what is the speed of the truck relative to the highway?

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? (Hint: See Problem \(34.69 . )(\mathrm{d})\) 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?

A thick-walled wine goblet sitting on a table can be considered to be a hollow glass sphere with an outer radius of 4.00 cm and an inner radius of 3.40 \(\mathrm{cm}\) . The index of refraction of the goblet glass is \(1.50 .(a)\) A beam of parallel light rays enters the side of the empty goblet along a horizontal radius. Where, if anywhere, will an image be formed? ( \(b\) ) The goblet is filled with white wine \((n=1.37)\) . Where is the image formed?

The Yerkes refracting telescope of the University of Chicago has an objective 1.02 \(\mathrm{m}\) in diameter with an \(f\) -number of \(19.0 .\) (This is the largest-diameter refracting telescope in the world.) What is its focal length?

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