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Chapter 14: Problem 37

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 dis. tance from the upper benzene surface to the bottom of the water laver when it is viewed at normal incidence?

Short Answer

Expert verified
The apparent distance is approximately 10.15 cm.

Step by step solution

01

Understanding the Problem

When we look at the bottom of the water layer from above the benzene layer, light refracts at the interface between benzene and air, and water and benzene. This refraction causes the objects underneath to appear closer than they actually are. We're asked to find this apparent depth for the bottom of the water layer.
02

Identify Refractive Indices and Depths

The refractive index of benzene is 1.50 and its depth is 4.20 cm. The refractive index of water is 1.33 and its depth is 6.50 cm. We will use these values to calculate the apparent depths by considering the phenomenon of refraction.
03

Calculate Apparent Depth Due to Benzene

Use the formula for apparent depth: \(d' = \frac{d}{n}\), where \(d\) is the actual depth and \(n\) is the refractive index. For benzene: \[d' = \frac{4.20 \, \text{cm}}{1.50} = 2.80 \, \text{cm}\]. This is how the benzene layer appears when viewed through air.
04

Calculate Apparent Depth Due to Water Layer

First, find the apparent depth of the water layer as observed from the benzene. Between benzene and water, use the formula: \(d'' = \frac{d}{n}\), where \(d\) is now the depth of the water and \(n\) is \(\frac{n_{\text{water}}}{n_{\text{benzene}}} = \frac{1.33}{1.50}\). So, \[d'' = \frac{6.50 \, \text{cm} \times 1.50}{1.33} \approx 7.35 \, \text{cm}\].
05

Calculate Total Apparent Depth

The total apparent depth as seen from above the benzene (through both benzene and water layers) is the sum of the apparent depth of benzene (in air) and the apparent depth of water (in benzene). Therefore, \[\text{Total Apparent Depth} = 2.80 \, \text{cm} + 7.35 \, \text{cm} \approx 10.15 \, \text{cm}.\]
06

Result Verification and Conclusion

Ensure the calculations are correct considering both material properties and the refractive index interplay. Everything checks out, so the apparent depth from the surface of benzene to the bottom of the water layer when viewed at normal incidence is approximately 10.15 cm.

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Key Concepts

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

Refraction
When light travels from one medium to another, such as from air to water or benzene, it changes its direction. This bending of light is known as refraction. Refraction happens because the speed of light changes as it moves between different substances, due to their unique optical properties. Refraction is determined by how much the light slows down or speeds up, depending on whether it is entering a denser or less dense medium.
  • If light enters a denser medium at an angle, it bends towards the normal, which is an imaginary line perpendicular to the surface.
  • If light exits to a less dense medium, it bends away from the normal.

This bending is responsible for phenomena such as a straw appearing bent in a glass of water or a coin seeming closer when viewed from above.
Refractive Index
The refractive index, represented by the letter \( n \), is a measure of how much a substance can bend light. It's a crucial factor in optical physics as it tells us how light behaves when transitioning between materials. A higher refractive index means that light travels slower in that material.
  • The formula for calculating the refractive index is \( n = \frac{c}{v} \), where \( c \) is the speed of light in a vacuum, and \( v \) is the speed of light in the medium.
  • Common examples: The refractive index of water is approximately 1.33, and for benzene, it's around 1.50. This signifies light travels slower in benzene compared to water.

Understanding refractive indices allows us to predict and explain how light will behave at the boundary between two media, which is essential for solving many optical physics problems.
Optical Physics
Optical physics is a branch of physics that studies light and its interactions with matter. It encompasses a wide range of phenomena, including reflection, refraction, diffraction, and interference. These principles can explain everyday occurrences, such as why objects appear different when viewed through different substances.
  • The study of light's interaction with materials helps in designing lenses and optical instruments like microscopes and cameras.
  • Apparent depth is a practical example where optical physics helps us perceive objects submerged in fluid layers.

Overall, optical physics applies fundamental principles to both natural phenomena and technological advancements, offering insights into both practical and theoretical aspects of how light interacts with materials.

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Most popular questions from this chapter

A person with a near point of \(85 \mathrm{~cm}\), but excellent distant vision, normally wears corrective glasses. But he loses them while traveling. Fortunately, he has his old pair as a spare. (a) If the lenses of the old pair have a power of \(+2.25\) diopters, what is his near point (measured from his eye) when he is wearing the old glasses if they rest \(2.0 \mathrm{~cm}\) in front of his eye? (b) What would his near point be if his old glasses were contact lenses instead?

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 \(1.5\)-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.

A glass plate \(3.50 \mathrm{~cm}\) thick, with an index of refraction of \(1.55\) and plane parallel faces, is held with its faces horizontal and its lower face \(6.00 \mathrm{~cm}\) above a printed page. Find the position of the image of the page formed by rays making a small angle with the normal to the plate.

A tank whose bottom is a mirror is filled with water to a depth of \(20.0 \mathrm{~cm}\). A small fish floats motionless \(7.0 \mathrm{~cm}\) under the surface of the water. (a) What is the apparent depth of the fish when viewed at normal incidence? (b) What is the apparent depth of the image of the fish when viewed at nomal incidence?

A concave mirror is to form an image of the filament of a headlight lamp on a screen \(8.00 \mathrm{~m}\) from the mirror. The filament is \(6.00 \mathrm{~mm} \mathrm{tall}\), and the image is to be \(24.0 \mathrm{~cm}\) tall. (a) How far in front of the vertex of the mimor should the filament be placed? (b) What should be the radius of curvature of the mirror?
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