/*! 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 74 \(\bullet$$\bullet\) A thick lay... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

\(\bullet$$\bullet\) A thick layer of oil is floating on the surface of water in a tank. A beam of light traveling in the oil is incident on the water interface at an angle of \(30.0^{\circ}\) from the normal. The refracted beam travels in the water at an angle of \(45.0^{\circ}\) from the normal. What is the refractive index of the oil?

Short Answer

Expert verified
The refractive index of the oil is approximately 1.88.

Step by step solution

01

Understand Snell's Law

Snell's Law is given by the equation \( n_1 \sin \theta_1 = n_2 \sin \theta_2 \), where \( n_1 \) and \( n_2 \) are the refractive indices of the media, and \( \theta_1 \) and \( \theta_2 \) are the angles of incidence and refraction, respectively. In this scenario, \( n_1 \) is the refractive index of oil, \( n_2 \) is the refractive index of water, \( \theta_1 = 30.0^{\circ} \) (angle in oil), and \( \theta_2 = 45.0^{\circ} \) (angle in water).
02

Refractive Index of Water

It's a well-known fact that the refractive index of water is approximately \( n_2 = 1.33 \). We will use this value in our calculations.
03

Rearrange Snell's Law for Oil's Refractive Index

We want to find the refractive index of oil \( n_1 \). Arrange Snell's Law: \( n_1 = \frac{n_2 \sin \theta_2}{\sin \theta_1} \). Substitute \( n_2 = 1.33 \), \( \theta_2 = 45.0^{\circ} \), and \( \theta_1 = 30.0^{\circ} \).
04

Calculate Sin Values

Calculate \( \sin 30.0^{\circ} \) and \( \sin 45.0^{\circ} \). We have \( \sin 30.0^{\circ} = 0.5 \) and \( \sin 45.0^{\circ} = \frac{\sqrt{2}}{2} \).
05

Plug in Values

Substitute everything into the equation: \( n_1 = \frac{1.33 \times \frac{\sqrt{2}}{2}}{0.5} \). Simplify this expression to get \( n_1 = 1.33 \times \sqrt{2} \).
06

Final Calculation

Calculate \( 1.33 \times \sqrt{2} \approx 1.33 \times 1.414 = 1.88 \). Thus, the refractive index of the oil is approximately \( 1.88 \).

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.

Refractive Index
The refractive index is an important property of materials in optics. It describes how light propagates through a medium. When light travels from one material to another, its speed changes, which bends the light path, a phenomenon known as refraction.

The refractive index, often denoted as \( n \), is defined as the ratio of the speed of light in a vacuum to the speed of light in the medium. Mathematically, this is given by:
  • \( n = \frac{c}{v} \)
where \( c \) is the speed of light in a vacuum, and \( v \) is the speed of light in the medium. The higher the refractive index, the more the light is bent when entering the material.

For practical uses, refractive indices are typically measured for visible light. Each material has a characteristic refractive index, which is crucial for designing optical devices, such as lenses and prisms.
Angles of Incidence and Refraction
The angles of incidence and refraction are key components in understanding light's behavior at the boundary between two media. When a light ray hits this boundary, the angle between the ray and an imaginary line perpendicular to the surface, called the "normal," is the angle of incidence.

Similarly, the angle between the refracted ray (the bent path) and the normal is called the angle of refraction. These angles are essential in applying Snell's Law, an equation that describes how light will refract based on the different refractive indices of the two media. Snell's Law is expressed as:
  • \( n_1 \sin \theta_1 = n_2 \sin \theta_2 \)
Here, \( n_1 \) and \( n_2 \) are the refractive indices of the initial and second media, respectively; \( \theta_1 \) is the angle of incidence, and \( \theta_2 \) is the angle of refraction.

Understanding these angles helps predict how and where light will travel when transitioning between substances such as air, water, or oil.
Optics
Optics is the branch of physics that deals with light and its interactions with matter. It encompasses a range of phenomena, including reflection, refraction, diffraction, and interference, all of which determine how light behaves and travels.

One key aspect of optics is understanding how light bends when it moves between different media, explained by Snell's Law. In designing optical equipment like glasses, cameras, and microscopes, the principles of optics ensure light is manipulated in the desired way to achieve clear images or specific effects.

Optical systems often use lenses and mirrors to direct or modify light. These elements rely heavily on refractive indices and proper calculation of angles of incidence and reflection/refraction to fulfill their functions. This makes optics crucial in various technological and scientific applications, ranging from everyday eyeglasses to advanced telescopic lenses for astronomy.

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

\(\bullet$$\bullet\) In a physics lab, light with wavelength 490 nm travels in air from a laser to a photocell in 17.0 ns. When a slab of glass 0.840 m thick is placed in the light beam, with the beam incident along the normal to the parallel faces of the slab, it takes the light 21.2 ns to travel from the laser to the photocell. What is the wavelength of the light in the glass?

If the entire slab (without the oil) is submerged in a fluid with an index of refraction of \(1.5,\) what will be the effect? A. The slab will appear to change color. B. Light striking the slab could be totally reflected. C. The slab will be very difficult to see. D. Light exiting the slab could be totally reflected.

\(\bullet\) Consider each of the electric- and magnetic-field orientations given next. In each case, what is the direction of propagation of the wave? (a) \(\vec{E}\) in the \(+x\) direction, \(\vec{B}\) in the \(+y\) direction. (b) \(\vec{E}\) in the \(-y\) direction, \(\vec{B}\) in the \(+x\) direction. (c) \(\vec{\boldsymbol{E}}\) in the \(+z\) direction, \(\vec{\boldsymbol{B}}\) in the \(-x\) direction. (d) \(\vec{\boldsymbol{E}}\) in the \(+y\) direction, \(\vec{\boldsymbol{B}}\) in the \(-z\) direction.

\(\bullet\) A glass plate having parallel faces and a refractive index of 1.58 lies at the bottom of a liquid of refractive index \(1.70 . \mathrm{A}\) ray of light in the liquid strikes the top of the glass at an angle of incidence of \(62.0^{\circ} .\) Compute the angle of refraction of this light in the glass.

\(\bullet$$\bullet\) Two sources of sinusoidal electromagnetic waves have average powers of 75 \(\mathrm{W}\) and 150 \(\mathrm{W}\) and emit uniformly in all directions. At the same distance from each source, what is the ratio of the maximum electric field for the 150 \(\mathrm{W}\) source to that of the 75 \(\mathrm{W}\) source?

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.