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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?

Short Answer

Expert verified
The wavelength of the light in the glass is approximately 64.73 nm.

Step by step solution

01

Understand the Problem

The problem involves calculating the wavelength of light in a medium (glass) given the time taken by light to travel a certain distance. We know that light travels with different speeds in different media, which affects its wavelength.
02

Calculate the Speed of Light in Glass

First, find out the speed of light in the glass. The light takes 21.2 ns to travel through the glass, which is 0.840 m thick. Thus, speed in glass, \( v_g = \frac{d}{t_g} = \frac{0.840 \, \text{m}}{21.2 \, \text{ns}} = 3.962 \times 10^7 \, \text{m/s} \).
03

Use Refractive Index

The refractive index \( n \) of the glass can be found using the ratio of the speed of light in vacuum (\( c = 3 \times 10^8 \, \text{m/s} \)) to the speed of light in glass \( v_g \). Thus, \( n = \frac{c}{v_g} = \frac{3 \times 10^8}{3.962 \times 10^7} \approx 7.57 \).
04

Calculate Wavelength in Glass

The wavelength of light in glass \( \lambda_g \) is related to its wavelength in air (\( \lambda_0 = 490 \, \text{nm} \)) by the refractive index, \( \lambda_g = \frac{\lambda_0}{n} = \frac{490 \times 10^{-9}\, \text{m}}{7.57} \approx 64.73 \, \text{nm} \).

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Key Concepts

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

Speed of Light
The speed of light, denoted by the symbol \(c\), is a fundamental constant in physics. In a vacuum, light travels at the speed of approximately \(3 \, \times \, 10^8\) meters per second. This speed is the fastest speed at which information or matter can travel, according to the theory of relativity.
However, the actual speed of light can vary when it passes through different materials, such as air, water, or glass. In a vacuum, light maintains its top speed, but in any other medium, it slows down.
Understanding how the speed of light changes in different media is crucial for solving many problems in optics, such as determining how lenses bend light or calculating the wavelength of light in materials.
Refractive Index
The refractive index, often represented by the symbol \(n\), is a dimensionless number that describes how light propagates through a medium. It is calculated by dividing the speed of light in a vacuum \((c)\) by the speed of light in the medium \((v)\).
The formula for the refractive index is:
\[ n = \frac{c}{v} \]
A higher refractive index indicates that light travels more slowly through the material. For example, glass has a higher refractive index than air, meaning light slows down more in glass than in air.
Refractive index is an essential concept in optics problem solving as it helps in understanding how light bends or refracts when it passes from one medium to another.
Light in Different Media
When light travels between different media, its speed and direction change. This phenomenon is known as refraction. Due to refraction, light will slow down or speed up, depending on the refractive indices of the media.
This change in speed affects not only the velocity but also the wavelength of the light in each medium.
When light enters a medium with a higher refractive index, its speed decreases, and the wavelength shortens. Conversely, in a medium with a lower refractive index, light speeds up, resulting in a longer wavelength.
Understanding how light behaves in different media helps us calculate the changes in light properties, crucial for designing lenses and optical devices.
Optics Problem Solving
Solving optics problems often involves applying multiple concepts such as speed, wavelength, and refractive index. These small yet crucial calculations are necessary to understand the behavior of light in various situations.
The problem-solving approach usually starts with identifying the necessary parameters like speed, distance, and time. Then, use formulas to find the unknowns, such as the refractive index or the new wavelength.
  • First, calculate the speed of light in the medium using distance and time measurements.
  • Then determine the refractive index by comparing it to the speed of light in a vacuum.
  • Finally, use the refractive index to find the new wavelength of light in the medium.
With these steps, you can systematically address and solve optics exercises, enhancing your understanding of how light interacts with materials.

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Most popular questions from this chapter

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 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?

Unpolarized light of intensity 20.0 \(\mathrm {W/cm}^2\) is incident on two polarizing filters. The axis of the first filter is at an angle of 25.0\(^\circ\) counterclockwise from the vertical (viewed in the direction the light is traveling), and the axis of the second filter is at 62.0\(^\circ\) counterclockwise from the vertical. What is the intensity of the light after it has passed through the second polarizer?

(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?

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?

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