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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, 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: n1=1.70 for the first glass and n2=1.58 for the second glass. The angle of incidence θ1 is given as 62.0.
02

Use Snell's Law to Set Up the Equation

Snell's Law states that n1sin(θ1)=n2sin(θ2), where θ2 is the angle of refraction. We need to compute θ2.
03

Substitute the Known Values

Substitute the known values into Snell's Law: 1.70sin(62.0)=1.58sin(θ2).
04

Compute sin(θ2)

Calculate sin(62.0) which is approximately 0.8829. Then set up the equation: 1.70×0.8829=1.58sin(θ2).
05

Solve for sin(θ2)

Rearrange the equation to solve for sin(θ2): sin(θ2)=1.70×0.88291.58. Compute this value to find sin(θ2)0.9508.
06

Compute the Angle of Refraction

To find θ2, take the inverse sine (arcsin) of sin(θ2). Thus, θ2=arcsin(0.9508). Using a calculator, θ272.0.

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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=cv
Where:
  • n is the refractive index
  • c is the speed of light in vacuum (approximately 3×108 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 θ2 using the equation:
  • n1sin(θ1)=n2sin(θ2)
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(θ2) by substituting known values, we found θ272.0, 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 flat piece of glass covers the top of a vertical cylinder that is completely filled with water. If a ray of light traveling in the glass is incident on the interface with the water at an angle of θa=36.2, the ray refracted into the water makes an angle of 49.8 with the normal to the interface. What is the smallest value of the incident angle θa for which none of the ray refracts into the water?

A light beam travels at 1.94×108 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 beam of light is traveling inside a solid glass cube that has index of refraction 1.62. It strikes the surface of the cube from the inside. (a) If the cube is in air, at what minimum angle with the normal inside the glass will this light not enter the air at this surface? (b) What would be the minimum angle in part (a) if the cube were immersed in water?

The refractive index of a certain glass is 1.66. For what incident angle is light reflected from the surface of this glass completely polarized if the glass is immersed in (a) air and (b) water?

Unpolarized light of intensity 20.0 W/cm2 is incident on two polarizing filters. The axis of the first filter is at an angle of 25.0 counterclockwise from the vertical (viewed in the direction the light is traveling), and the axis of the second filter is at 62.0 counterclockwise from the vertical. What is the intensity of the light after it has passed through the second polarizer?

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