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Chapter 34: Problem 70

A layer of benzene \((n=1.50)\) that is \(4.20 \mathrm{~cm}\) deep floats on water \((n=1.33)\) that is \(5.70 \mathrm{~cm}\) deep. What is the apparent distance from the upper benzene surface to the bottom of the water when you view these layers at normal incidence?

Short Answer

Expert verified
The apparent distance from the upper benzene surface to the bottom of the water when viewed at normal incidence is \(7.09 \mathrm{~cm}\).

Step by step solution

01

Calculate Apparent Depth for Benzene

First we will find the apparent depth for the Benzene layer. The formula for apparent depth (\(d_a\)) is actual depth (\(d\)) divided by the refractive index (\(n\)). Thus, \(d_a = d/n\). Substituting the values, \(d_a = 4.20/1.50 = 2.80 \mathrm{~cm}\).
02

Calculate Apparent Depth for Water

The same formula can be used for the water layer. Thus, \(d_a = 5.70/1.33 = 4.29 \mathrm{~cm}\).
03

Calculate Total Apparent Distance

The total apparent distance is the apparent depth of the benzene layer plus the apparent depth of the water layer. Hence, the total apparent distance is \(2.80 + 4.29 = 7.09 \mathrm{~cm}\).

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Key Concepts

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

Refractive Index
The refractive index, often symbolized as \( n \), is a measure of how much light slows down as it passes through a material. It is a dimensionless number that describes how light travels in different media. Simply put, the higher the refractive index of a material, the slower light moves through it compared to its speed in a vacuum.
  • A vacuum has a refractive index of 1, which is the baseline reference.
  • Common materials such as glass, water, and benzene have refractive indices greater than 1.
  • For example, the refractive index of benzene is 1.50, indicating light travels slower in benzene than in the air.
Understanding the refractive index is crucial in optics as it impacts how light behaves, especially in phenomena like refraction, where the direction of light changes as it moves between different media.
Apparent Depth
Apparent depth is a phenomenon where an object submerged in a medium, such as a liquid, appears to be closer to the surface than it actually is. This occurs due to the bending of light, which changes the perceived depth of the object. The formula for apparent depth is given by dividing the actual depth by the refractive index of the medium:\[ d_a = \frac{d}{n} \]
  • In the context of the exercise, benzene and water have different refractive indices, so light bends differently in each layer.
  • Thus, a typical observer sees the bottom of the benzene layer at an apparent depth of 2.80 cm rather than its actual 4.20 cm.
  • Similarly, the bottom of the water layer appears at 4.29 cm instead of 5.70 cm.
This concept plays a critical role in designing systems like lenses and water tanks, where knowing the true position of objects is necessary.
Normal Incidence
Normal incidence refers to the condition where light rays hit a surface directly at 90 degrees. When light encounters a boundary under normal incidence, there is no angular deviation. In simpler terms, light travels straight through, without bending, as long as each layer is considered uniform and parallel.
  • This exercise assumes normal incidence, meaning the light travels perpendicularly through both the benzene and water layers.
  • Thus, the only factor in the perceived depth is the change in speed of light due to differing refractive indices, not the angle of entry.
This concept is less common in real-world scenarios, where light often strikes at various angles, causing more complex refraction effects.
Light Refraction
Refraction is the bending of light as it passes from one transparent medium to another. This bending occurs due to a change in light's speed and direction caused by different refractive indices in each medium.
  • For instance, when light travels from benzene (n=1.50) into water (n=1.33), it speeds up, causing it to bend at the interface of these two layers.
  • Refraction is responsible for the apparent depth effect discussed earlier.
  • Applications of refraction are vast, ranging from corrective lenses in eyeglasses to the intricate crafting of camera lenses.
Understanding how refraction works helps explain everyday observations, like why a pencil looks bent when placed in a glass of water.

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Most popular questions from this chapter

Three thin lenses, each with a focal length of \(40.0 \mathrm{~cm},\) are aligned on a common axis; adjacent lenses are separated by \(52.0 \mathrm{~cm} .\) Find the position of the image of a small object on the axis, \(80.0 \mathrm{~cm}\) to the left of the first lens.

You hold a spherical salad bowl \(60 \mathrm{~cm}\) in front of your face with the bottom of the bowl facing you. The bowl is made of polished metal with a \(35 \mathrm{~cm}\) radius of curvature. (a) Where is the image of your \(5.0-\mathrm{cm}\) -tall nose located? (b) What are the image's size, orientation, and nature (real or virtual)?

A coin is placed next to the convex side of a thin spherical glass shell having a radius of curvature of \(18.0 \mathrm{~cm} .\) Reflection from the surface of the shell forms an image of the \(1.5-\mathrm{cm}\) -tall coin that is \(6.00 \mathrm{~cm}\) behind the glass shell. Where is the coin located? Determine the size, orientation, and nature (real or virtual) of the image.

To determine whether a frog can judge distance by means of the amount its lens must move to focus on an object, researchers covered one eye with an opaque material. An insect was placed in front of the frog, and the distance that the frog snapped its tongue out to catch the insect was measured with high-speed video. The experiment was repeated with a contact lens over the eye to determine whether the frog could correctly judge the distance under these conditions. If such an experiment is performed twice, once with a lens of power \(-9 \mathrm{D}\) and once with a lens of power \(-15 \mathrm{D},\) in which case does the frog have to focus at a shorter distance, and why? (a) With the \(-9 \mathrm{D}\) lens; because the lenses are diverging, the lens with the longer focal length creates an image that is closer to the frog. (b) With the \(-15 \mathrm{D}\) lens; because the lenses are diverging, the lens with the shorter focal length creates an image that is closer to the frog. (c) With the \(-9 \mathrm{D}\) lens; because the lenses are converging, the lens with the longer focal length creates a larger real image. (d) With the -15 D lens; because the lenses are converging, the lens with the shorter focal length creates a larger real image.

An object \(0.600 \mathrm{~cm}\) tall is placed \(24.0 \mathrm{~cm}\) to the left of the vertex of a concave spherical mirror. The image of the object is inverted and is \(2.50 \mathrm{~cm}\) tall. What is the radius of curvature of the mirror?
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