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A ray of light is incident on a plane surface separating two sheets of glass with refractive indexes 1.70 and 1.58. The angle of incidence is 62.0\(^\circ\), and the ray originates in the glass with \(n\) = 1.70. Compute the angle of refraction.

Short Answer

Expert verified
The angle of refraction is approximately 72.0°.

Step by step solution

01

Identify the Known Values

We have the refractive indexes of the two sheets of glass: \( n_1 = 1.70 \) for the first glass and \( n_2 = 1.58 \) for the second glass. The angle of incidence \( \theta_1 \) is given as 62.0\(^\circ\).
02

Use Snell's Law to Set Up the Equation

Snell's Law states that \( n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \), where \( \theta_2 \) is the angle of refraction. We need to compute \( \theta_2 \).
03

Substitute the Known Values

Substitute the known values into Snell's Law: \( 1.70 \sin(62.0^\circ) = 1.58 \sin(\theta_2) \).
04

Compute \( \sin(\theta_2) \)

Calculate \( \sin(62.0^\circ) \) which is approximately 0.8829. Then set up the equation: \( 1.70 \times 0.8829 = 1.58 \sin(\theta_2) \).
05

Solve for \( \sin(\theta_2) \)

Rearrange the equation to solve for \( \sin(\theta_2) \): \( \sin(\theta_2) = \frac{1.70 \times 0.8829}{1.58} \). Compute this value to find \( \sin(\theta_2) \approx 0.9508 \).
06

Compute the Angle of Refraction

To find \( \theta_2 \), take the inverse sine (arcsin) of \( \sin(\theta_2) \). Thus, \( \theta_2 = \arcsin(0.9508) \). Using a calculator, \( \theta_2 \approx 72.0^\circ \).

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Key Concepts

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

Refractive Index
Refractive index is a measure of how much light or any other radiation is bent, or refracted, when it enters a material. Each material has a unique refractive index, denoted as 'n'. This index is a dimensionless number that describes how light propagates through the medium compared to the vacuum of space. Without this concept, understanding how lenses work would be challenging.

The formula for refractive index is derived from the ratio of the speed of light in a vacuum to the speed of light in the given medium. Mathematically, it is expressed as:
  • \( n = \frac{c}{v} \)
Where:
  • \( n \) is the refractive index
  • \( c \) is the speed of light in vacuum (approximately \( 3 \times 10^8 \) m/s)
  • \( v \) is the speed of light in the medium
Understanding refractive index is crucial in optics, as it affects lens design, the focus of images, and the visual quality of optical instruments like glasses and cameras.
Angle of Incidence
The angle of incidence is the angle between the incident ray and the normal (an imaginary line perpendicular to the surface at the point of incidence). In optics, this concept is vital because it determines how light behaves when it strikes a surface.

When light travels from one medium to another with a different refractive index, the angle of incidence can greatly influence the degree to which the light is bent, or refracted. This concept is used in Snell's Law to predict and calculate the path of light.

The angle of incidence is crucial for designing optical devices, as it helps to control how light is to be redirected. For example, in our exercise, the angle of incidence was 62.0 degrees, which helps us in determining how the light will bend as it passes from one glass to another.
Angle of Refraction
The angle of refraction is the angle between the refracted ray and the normal at the point of refraction. Like the angle of incidence, it is key in understanding how light moves through different media. Snell's Law links the angle of incidence and the angle of refraction through their respective refractive indices.

In our given exercise, Snell's Law was used to calculate the angle of refraction \( \theta_2 \) using the equation:
  • \( n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \)
Understanding the angle of refraction is fundamental to designing scientific instruments and technologies that require precision, such as telescopes, microscopes, and corrective lenses. In the exercise, once we solved for \( \sin(\theta_2) \) by substituting known values, we found \( \theta_2 \approx 72.0^\circ \), illustrating how light deviates as it transitions between materials with different refractive indices.
Optics
Optics is the branch of physics that deals with the study of light and its interactions with different materials. It explains phenomena such as reflection, refraction, and dispersion which are pivotal in our understanding of visual perception and the creation of various optical technologies.

In the world of optics, Snell's Law is a fundamental concept used to describe how light waves change direction when they pass from one medium into another. This law is crucial for calculating how lenses will focus light and how devices like glasses or contact lenses can correct vision.

Optics not only covers theoretical aspects but is applied in various fields, including designing cameras, telescopes, microscopes, and even fiber optics for communication technologies. The practical applications of optics in everyday life make it an essential topic for study. By understanding concepts like refractive index and angles of incidence and refraction, we become better equipped to innovate and deploy technologies that rely on the manipulation of light.

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Most popular questions from this chapter

A thin layer of ice (\(n\) = 1.309) floats on the surface of water (\(n\) = 1.333) in a bucket. A ray of light from the bottom of the bucket travels upward through the water. (a) What is the largest angle with respect to the normal that the ray can make at the ice-water interface and still pass out into the air above the ice? (b) What is this angle after the ice melts?

Unpolarized light with intensity \(I_0\) is incident on two polarizing filters. The axis of the first filter makes an angle of 60.0\(^\circ\) with the vertical, and the axis of the second filter is horizontal. What is the intensity of the light after it has passed through the second filter?

A light beam travels at \(1.94 \times 10^8\) m/s in quartz. The wavelength of the light in quartz is 355 nm. (a) What is the index of refraction of quartz at this wavelength? (b) If this same light travels through air, what is its wavelength there?

(a) A tank containing methanol has walls 2.50 cm thick made of glass of refractive index 1.550. Light from the outside air strikes the glass at a 41.3\(^\circ\) angle with the normal to the glass. Find the angle the light makes with the normal in the methanol. (b) The tank is emptied and refilled with an unknown liquid. If light incident at the same angle as in part (a) enters the liquid in the tank at an angle of 20.2\(^\circ\) from the normal, what is the refractive index of the unknown liquid?

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?

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