/*! 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 70 A layer of benzene \((n=1.50) 2.... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A layer of benzene \((n=1.50) 2.60 \mathrm{cm}\) deep floats on water \((n-1.33)\) that is 6.50 \(\mathrm{cm}\) deep. What is the apparent distance from the upper benzene surface to the bottom of the water layer when it is viewed at normal incidence?

Short Answer

Expert verified
The apparent distance from the benzene surface to the bottom of the water is approximately 6.62 cm.

Step by step solution

01

Understanding Refraction Index

The refractive index ( ) of a medium is a measure of how much light is bent, or refracted, when entering a material. Benzene has a refractive index of 1.50, and water has an index of 1.33. These values mean light travels slower in these materials compared to a vacuum.
02

Depth Correction for Benzene Layer

To find the apparent depth of the benzene layer, we apply the formula for apparent depth: \( \text{Apparent depth} = \frac{\text{Actual depth}}{n} \). For benzene, the actual depth is 2.60 cm and the refractive index is 1.50. Thus, \( \text{Apparent depth of benzene} = \frac{2.60}{1.50} \approx 1.73 \) cm.
03

Depth Correction for Water Layer

Next, we calculate the apparent depth of the water layer using the same formula. The actual depth of the water is 6.50 cm, and the refractive index is 1.33. Thus, \( \text{Apparent depth of water} = \frac{6.50}{1.33} \approx 4.89 \) cm.
04

Calculate Total Apparent Depth

The total apparent distance from the upper surface of the benzene to the bottom of the water layer is the sum of the apparent depths of benzene and water. Therefore, \( \text{Total apparent depth} = 1.73 + 4.89 = 6.62 \) cm.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Refraction Index
The refraction index, also known as the refractive index, is a fundamental concept in optics that describes how light propagates through different media. It is represented by the symbol \( n \) and is defined as the ratio of the speed of light in a vacuum to the speed of light in the medium. This means that if the refractive index of a material is high, light travels slower through that material compared to a vacuum.
A refractive index value greater than 1 indicates that the medium can bend light to a certain extent. For instance, in the given example, benzene has a refractive index of 1.50, and water has a refractive index of 1.33. These values show how much light slows down and bends when passing through benzene and water.
In simpler terms, the refractive index tells us how much the path of light is altered when passing from one transparent medium to another. Different materials have different refractive indices, which leads to unique bending properties of light when it moves from air into those substances.
Apparent Depth
Apparent depth is a concept related to the way light moves through materials with different refractive indices. It explains why objects under a surface may appear closer than they actually are. This illusion is caused by the bending of light as it passes through mediums with different refractive indices.
In the context of this exercise, we have two layers: benzene and water. Due to their refractive indices, the actual depth of these layers appears to be different when viewed from above. The apparent depth can be calculated using the formula: \[ \text{Apparent depth} = \frac{\text{Actual depth}}{n} \]where \( n \) is the refractive index of the medium.
For benzene with an actual depth of 2.60 cm and \( n = 1.50 \), the apparent depth comes out to be approximately 1.73 cm. Similarly, for water with an actual depth of 6.50 cm and \( n = 1.33 \), its apparent depth is approximately 4.89 cm. By summing these apparent depths, you get the complete apparent depth from the top surface of benzene to the bottom of the water layer, which results in 6.62 cm.
Light Refraction
Light refraction is the bending or change in direction of light as it passes from one medium to another. This change in direction is due to the change in speed as light moves through various materials. The amount of bending depends on the refractive index of the two media involved.
Refraction occurs because light travels at different speeds in different materials. For example, it travels fastest in a vacuum and slower in denser materials like water or glass. This change in speed results in light bending at the interface between two materials.
Some everyday experiences of refraction include the bending of a straw placed in a glass of water or the shimmering heat effect on a road in the distance on a hot day. In such cases, light is bending because of the differences in refractive indices between air and water, or due to temperature gradients in air respectively.
  • Refraction is responsible for phenomena like apparent depth; objects under water appear nearer than they actually are.
  • Knowing the refractive indices helps us calculate exactly how light will behave when passing from one medium to another.
  • This principle is extensively used in lens design, cameras, glasses, and other optical devices.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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?

Three-Dimensional Image. The longitudinal magnification is defined as \(m^{\prime}=d s^{\prime} / d s .\) It relates the longitudinal dimension of a small object to the longitudinal dimension of its image. (a) Show that for a spherical mirror, \(m^{\prime}=-m^{2}\) . What is the significance of the fact that \(m^{\prime}\) is always negative? (b) A wire frame in the form of a small cube 1.00 \(\mathrm{mm}\) on a side is placed with its center on the axis of a concave mirror with radius of curvature 150.0 \(\mathrm{cm}\) . The sides of the cube are all either parallel or perpendicular to the axis. The cube face toward the mirror is 200.0 \(\mathrm{cm}\) to the left of the mirror vertex. Find (i) the location of the image of this face and of the opposite face of the cube; (ii) the lateral and longitudinal magnifications; (iii) the shape and dimensions of each of the six faces of the image.

A converging lens with a focal length of 12.0 \(\mathrm{cm}\) forms a virtual image 8.00 \(\mathrm{mm}\) tall, 17.0 \(\mathrm{cm}\) to the right of the lens. Determine the position and size of the object. Is the image erect or inverted? Are the object and image on the same side or opposite sides of the lens? Draw a principal-ray diagram for this situation.

A certain microscope is provided with objectives that have focal lengths of \(16 \mathrm{mm}, 4 \mathrm{mm},\) and 1.9 \(\mathrm{mm}\) and with eyepieces that have angular magnifications of \(5 \times\) and \(10 \times .\) Each objective forms an image 120 \(\mathrm{mm}\) beyond its second focal point. Determine (a) the largest overall angular magnification obtainable and (b) the least overall angular magnification obtainable.

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 \(\mathrm{mm}\) tall, and the image is to be 36.0 \(\mathrm{cm}\) tall. (a) How far in front of the vertex of the mirror should the filament be placed? (b) What should be the radius of curvature of the mirror?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.