/*! 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 10 A layer of oil \((n=1.45)\) floa... [FREE SOLUTION] | 91Ó°ÊÓ

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A layer of oil \((n=1.45)\) floats on an unknown liquid. A ray of light originates in the oil and passes into the unknown liquid. The angle of incidence is \(64.0^{\circ},\) and the angle of refraction is \(53.0^{\circ} .\) What is the index of refraction of the unknown liquid?

Short Answer

Expert verified
The index of refraction of the unknown liquid is approximately 1.635.

Step by step solution

01

Identify the Known Values

The index of refraction of the oil is given as \(n_1 = 1.45\). The angle of incidence \(\theta_1\) is \(64.0^{\circ}\), and the angle of refraction \(\theta_2\) is \(53.0^{\circ}\). We need to find the index of refraction of the unknown liquid, \(n_2\).
02

Apply Snell's Law

Snell's Law relates the indices of refraction and the angles of incidence and refraction as follows: \[ n_1 \sin \theta_1 = n_2 \sin \theta_2 \]where \(n_1\) and \(\theta_1\) are for the oil, and \(n_2\) and \(\theta_2\) are for the unknown liquid. We can rearrange this equation to solve for \(n_2\):\[ n_2 = \frac{n_1 \sin \theta_1}{\sin \theta_2} \]
03

Calculate the Sine Values

Calculate \(\sin \theta_1\) and \(\sin \theta_2\):- \(\sin 64.0^{\circ} \approx 0.8988\)- \(\sin 53.0^{\circ} \approx 0.7986\)
04

Solve for the Index of Refraction

Substitute the known values and sine calculations into the Snell's Law equation to find \(n_2\):\[ n_2 = \frac{1.45 \times 0.8988}{0.7986} \approx 1.635 \]

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Key Concepts

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

Optics
Optics is the fascinating branch of physics that studies light and its interaction with various materials. This field examines how light behaves when it encounters different surfaces and substances, such as air, water, or glass. Things like reflection, refraction, and dispersion are all important aspects of optics. In the case of refraction, for example, optics seeks to understand how light bends when it transitions between different media. This bending effect is what allows us to see through lenses and everyday objects like eyeglasses and cameras. Understanding optics can help us not only in everyday life but also in technological innovations like fiber optics and advanced imaging systems.
In the scenario of the exercise, optics is applied by analyzing the light as it moves from the oil into the unknown liquid, allowing us to understand how it bends at the interface.
Index of Refraction
The index of refraction is a unique property of materials that indicates how much they slow down light. When light enters a new medium, it doesn't travel at its original speed; instead, its speed changes depending on the material. The index of refraction, usually denoted by the symbol \(n\), is a ratio of the speed of light in a vacuum to the speed of light in the medium. For instance, the index of refraction for oil given in the problem is \(n_1 = 1.45\), meaning light travels slower in oil than in a vacuum. Typically, if the index of refraction is greater than 1, light travels slower in that medium than in a vacuum.
In the exercise, the challenge is to find the index of refraction for the unknown liquid. By using Snell's Law, one can calculate this unknown index based on the given indices and angles of the known media.
Angle of Incidence
The angle of incidence refers to the angle between the incoming light ray and the normal (an imaginary line perpendicular) to the surface at the point of contact. It is important in determining how light will behave as it moves from one medium to another. This angle directly influences refraction according to Snell's Law. Imagine shining a flashlight onto a smooth surface; the angle at which the light hits the surface is the angle of incidence.
In the given exercise, the angle of incidence for the ray within the oil is \(64.0^{\circ}\). This starting angle is essential for calculating how the light will bend as it enters the unknown liquid.
Angle of Refraction
When light passes from one medium into another, the angle of refraction describes the angle between the refracted (bent) ray and the normal line to the surface in the second medium. This angle is crucial in determining how much a light beam will deviate from its original path, guided by Snell's Law. Light bends differently depending on the indices of refraction of the two media involved. The greater the difference, the more pronounced the bending can be.
In the problem, the angle of refraction as the light passes into the unknown liquid is \(53.0^{\circ}\). This information, along with the angle of incidence and known refractive indices, helps determine the index of refraction for the unknown substance using Snell's Law. This demonstrates the practical application of understanding the angle of refraction in optics.

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Most popular questions from this chapter

Two converging lenses are separated by \(24.00 \mathrm{cm} .\) The focal length of each lens is \(12.00 \mathrm{cm} .\) An object is placed \(36.00 \mathrm{cm}\) to the left of the lens that is on the left. Determine the final image distance relative to the lens on the right.

In a compound microscope, the focal length of the objective is \(3.50 \mathrm{cm}\) and that of the eyepiece is \(6.50 \mathrm{cm}\). The distance between the lenses is \(26.0 \mathrm{cm} .\) (a) What is the angular magnification of the microscope if the person using it has a near point of \(35.0 \mathrm{cm} ?\) (b) If, as usual, the first image lies just inside the focal point of the eyepiece (see Figure 26.32 ), how far is the object from the objective? (c) What is the magnification (not the angular magnification) of the objective?

A spectator, seated in the left-field stands, is watching a baseball player who is \(1.9 \mathrm{m}\) tall and is \(75 \mathrm{m}\) away. On a \(\mathrm{TV}\) screen, located \(3.0 \mathrm{m}\) from a person watching the game at home, the image of this same player is \(0.12 \mathrm{m}\) tall. Find the angular size of the player as seen by (a) the spectator watching the game live and (b) the TV viewer. (c) To whom does the player appear to be larger?

A farsighted woman breaks her current eyeglasses and is using an old pair whose refractive power is 1.660 diopters. since these eyeglasses do not completely correct her vision, she must hold a newspaper \(42.00 \mathrm{cm}\) from her eyes in order to read it. She wears the eyeglasses \(2.00 \mathrm{cm}\) from her eyes. How far is her near point from her eyes?

A diverging lens \((f=-10.0 \mathrm{cm})\) is located \(20.0 \mathrm{cm}\) to the left of a converging lens \((f=30.0 \mathrm{cm}) .\) A \(3.00-\mathrm{cm}-\) tall object stands to the left of the diverging lens, exactly at its focal point. (a) Determine the distance of the final image relative to the converging lens. (b) What is the height of the final image (including the proper algebraic sign)?

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