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Light traveling in air is incident on the surface of a block of plastic at an angle of 62.7\(^\circ\) to the normal and is bent so that it makes a 48.1\(^\circ\) angle with the normal in the plastic. Find the speed of light in the plastic.

Short Answer

Expert verified
The speed of light in the plastic is approximately \(2.31 \times 10^8 \ m/s\).

Step by step solution

01

Identify Given Values

We are given the angle of incidence \( \theta_1 = 62.7^\circ \) and the angle of refraction \( \theta_2 = 48.1^\circ \).
02

Apply Snell's Law

Snell's law states that \( n_1 \sin \theta_1 = n_2 \sin \theta_2 \), where \( n_1 \) is the refractive index of air (approximately 1) and \( n_2 \) is the refractive index of the plastic that we need to find.
03

Calculate Refractive Index of Plastic

Since \( n_1 \approx 1 \), we have \( \sin(62.7^\circ) = n_2 \sin(48.1^\circ) \). Solving for \( n_2 \), we have \( n_2 = \frac{ \sin(62.7^\circ) }{ \sin(48.1^\circ) } \approx 1.296 \).
04

Use the Refractive Index to Find the Speed of Light

The speed of light in a medium is given by \( v = \frac{c}{n} \), where \( c \approx 3 \times 10^8 \ m/s \) is the speed of light in a vacuum and \( n = 1.296 \) is the refractive index of the plastic.
05

Calculate the Speed of Light in the Plastic

Plugging the values into the equation, we get: \( v = \frac{3 \times 10^8 \ m/s}{1.296} \approx 2.31 \times 10^8 \ m/s \).

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Key Concepts

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

Angle of Incidence
The angle of incidence is a fundamental concept when studying how light interacts with different media. It refers to the angle at which a light ray approaches a surface, measured from the normal, which is an imaginary line perpendicular to the surface. For instance, in the given exercise, light hits the surface of the plastic block at an angle of 62.7 degrees from the normal. Understanding this angle helps us predict how the light will behave as it passes into the new medium.
To visualize this, imagine a beam of light striking the surface of a clear plastic sheet. The line where this light meets the plastic forms the angle of incidence. Properly identifying this angle is crucial for applying Snell's Law, which relates the angles of incidence and refraction in different media.
Angle of Refraction
The angle of refraction is the angle that a light ray forms as it exits one medium and enters another. This angle is also measured relative to the normal. In our exercise, after the light enters the plastic, it bends and forms a 48.1-degree angle with the normal.
The process of bending, or refraction, occurs because light travels at different speeds in different media. As light enters a denser medium from a less dense medium, it slows down, causing it to change direction. This alteration of angle is what we observe as refraction. The angle of refraction is key to understanding how lenses work, how prisms disperse light, and much more in optics.
  • Occurs due to the speed change of light.
  • Helps to determine the extent of light bending.
Refractive Index
The refractive index is a measure that describes how fast light travels through a material compared to the speed of light in a vacuum. It is a crucial element of Snell's Law, which is expressed as \( n_1 \sin \theta_1 = n_2 \sin \theta_2 \). In our case, the air (n鈧) has a refractive index of approximately 1, while the plastic's refractive index (n鈧) is calculated to be about 1.296.
This index helps us determine how much a ray of light will bend, depending on the media it is traveling through. If light moves from a medium with a lower refractive index like air to a higher one like plastic, it bends towards the normal.
  • Higher index = stronger bending effect.
  • Important for designing lenses and fiber optics.
Speed of Light
The speed of light in a vacuum is a constant \( c \) of approximately \( 3 \times 10^8 \ m/s \). However, when light travels through different materials, its speed changes, which is integral to understanding refractive behavior. In the exercise, we find the speed of light in the plastic using the equation \( v = \frac{c}{n} \), where \( n \) is the refractive index of the plastic.
After calculating, we found that light travels at roughly \( 2.31 \times 10^8 \ m/s \) in the plastic. Knowing these speeds and how they vary between substances helps us understand phenomena such as refraction and reflection, and is essential for designing optical devices like cameras and glasses.
  • Speeds differ across materials due to density differences.
  • Crucial for high-precision scientific applications.

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Most popular questions from this chapter

Three polarizing filters are stacked, with the polarizing axis of the second and third filters at 23.0\(^\circ\) and 62.0\(^\circ\), respectively, to that of the first. If unpolarized light is incident on the stack, the light has intensity 55.0 \(\mathrm {W/cm}^2\) after it passes through the stack. If the incident intensity is kept constant but the second polarizer is removed, what is the intensity of the light after it has passed through the stack?

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.

Optical fibers are constructed with a cylindrical core surrounded by a sheath of cladding material. Common materials used are pure silica \(n_2 = 1.4502\) for the cladding and silica doped with germanium \(n_1 = 1.4652\) for the core. (a) What is the critical angle \(\theta_{crit}\) for light traveling in the core and reflecting at the interface with the cladding material? (b) The numerical aperture (NA) is defined as the angle of incidence \(theta_i\) at the flat end of the cable for which light is incident on the core-cladding interface at angle \(\theta_{crit}\) (\(\textbf{Fig. P33.46}\)). Show that sin \(\theta_i\) =\( \sqrt {n^2_1 - n^2_2}\) . (c) What is the value of \(\theta_i\) for \(n_1\) = 1.465 and \(n_2\) = 1.450?

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 critical angle for total internal reflection at a liquid air interface is 42.5\(^\circ\). (a) If a ray of light traveling in the liquid has an angle of incidence at the interface of 35.0\(^\circ\), what angle does the refracted ray in the air make with the normal? (b) If a ray of light traveling in air has an angle of incidence at the interface of 35.0\(^\circ\), what angle does the refracted ray in the liquid make with the normal?

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