/*! 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 1.23.

Step by step solution

01

Understand Snell's Law

Snell's Law relates the angles and indices of refraction of two media. It is given by the formula: \( n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \), where \( n_1 \) and \( n_2 \) are the indices of refraction of the first and second media, and \( \theta_1 \) and \( \theta_2 \) are the angles of incidence and refraction, respectively.
02

Identify Known Values

For this problem, the known values are: \( n_1 = 1.45 \) (the index of refraction of oil), \( \theta_1 = 64.0^{\circ} \) (the angle of incidence in the oil), and \( \theta_2 = 53.0^{\circ} \) (the angle of refraction in the unknown liquid).

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Key Concepts

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

Index of Refraction
The index of refraction, often symbolized by the letter \( n \), is a fundamental property of any transparent material. It measures how much the speed of light is reduced inside a medium compared to its speed in a vacuum. The higher the index, the slower light travels in that material. This concept is crucial when dealing with light traveling between different media. For example, light moves slower in glass or water than in air, hence their indices of refraction are greater than 1. The index of refraction can be denoted mathematically as:
  • \( n = \frac{c}{v} \)
where \( c \) is the speed of light in a vacuum and \( v \) is the speed of light in the medium.
In the given exercise, the index of refraction for oil was provided as \( n = 1.45 \). This implies that light travels at \( \frac{1}{1.45} \) times its speed in a vacuum when moving through the oil.
Angle of Incidence
The angle of incidence is the angle that a light beam makes with the normal (an imaginary line perpendicular to the surface) upon entering a new medium. Think of it like this: when you shine a flashlight straight down onto a table, the light beam is nearly zero degrees from the normal; move it sideways, and you increase the angle.
In the Snell's Law equation, represented as \( \theta_1 \), the angle of incidence demonstrates how light travels from one medium to another. It's pivotal to correctly measure this angle from the normal, not from the surface. In the exercise above, the angle of incidence is given as \( 64.0^{\circ} \), indicating the angle between the incoming light ray and the imaginary normal to the oil’s surface.
Angle of Refraction
When light enters a different medium, it bends; this bending is called refraction, and the angle is known as the angle of refraction. This angle, denoted as \( \theta_2 \) in Snell's Law, is the angle between the refracted ray and the normal in the second medium. It happens because light changes speed when moving between materials with different indices of refraction.
The angle of refraction is crucial in determining how much the light path bends. As per Snell’s Law, the relationship for refraction can be calculated by equating the product of the index of refraction and sine of the angle of incidence to the product of the index of refraction and sine of the angle of refraction for the new medium:
  • \( n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \)
In the given exercise, the angle of refraction was \( 53.0^{\circ} \), which allowed the calculation of the unknown liquid's index of refraction. This reveals how differently light behaves in comparison to the oil.

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Most popular questions from this chapter

An object is placed 20.0 \(\mathrm{cm}\) to the left of a diverging lens \((f=-8.00 \mathrm{cm})\) . A concave mirror \((f=12.0 \mathrm{cm})\) is placed 30.0 \(\mathrm{cm}\) to the right of the lens. (a) Find the final image distance, measured relative to the mirror. (b) Is the final image real or virtual? (c) Is the final image upright or inverted with respect to the original object?

A microscope for viewing blood cells has an objective with a focal length of 0.50 cm and an eyepiece with a focal length of 2.5 cm. The distance between the objective and eyepiece is 14.0 cm. If a blood cell subtends an angle of 2.1 \(\times 10^{-5}\) rad when viewed with the naked eye at a near point of \(25.0 \mathrm{cm},\) what angle (magnitude only) does it subtend when viewed through the microscope?

The near point of a naked eye is 32 cm. When an object is placed at the near point and viewed by the naked eye, it has an angular size of 0.060 rad. A magnifying glass has a focal length of 16 cm, and is held next to the eye. The enlarged image that is seen is located 64 cm from the magnifying glass. Determine the angular size of the image.

A stargazer has an astronomical telescope with an objective whose focal length is 180 cm and an eyepiece whose focal length is 1.20 cm. He wants to increase the angular magnification of a galaxy under view by replacing the telescope’s eyepiece. Once the eyepiece is replaced, the barrel of the telescope must be adjusted to bring the galaxy back into focus. If the barrel can only be shortened by 0.50 cm from its current length, what is the best angular magnification the stargazer will be able to achieve?

The near point of a naked eye is 25 cm. When placed at the near point and viewed by the naked eye, a tiny object would have an angular size of \(5.2 \times 10^{-5}\) rad. When viewed through a compound microscope, however, it has an angular size of \(-8.8 \times 10^{-3}\) rad. (The minus sign indicates that the image produced by the microscope is inverted.) The objective of the microscope has a focal length of 2.6 cm, and the distance between the objective and the eyepiece is 16 cm. Find the focal length of the eyepiece.

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