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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

The vitreous humor, a transparent, gelatinous fluid that fills most of the eyeball, has an index of refraction of 1.34. Visible light ranges in wavelength from 380 nm (violet) to 750 nm (red), as measured in air. This light travels through the vitreous humor and strikes the rods and cones at the surface of the retina. What are the ranges of (a) the wavelength, (b) the frequency, and (c) the speed of the light just as it approaches the retina within the vitreous humor?

When the sun is either rising or setting and appears to be just on the horizon, it is in fact below the horizon. The explanation for this seeming paradox is that light from the sun bends slightly when entering the earth’s atmosphere, as shown in Fig. \(\textbf{P33.51}\). Since our perception is based on the idea that light travels in straight lines, we perceive the light to be coming from an apparent position that is an angle \(\delta\) above the sun's true position. (a) Make the simplifying assumptions that the atmosphere has uniform density, and hence uniform index of refraction \(n\), and extends to a height \(h\) above the earth's surface, at which point it abruptly stops. Show that the angle \(\delta\) is given by $${ \delta = \mathrm{arcsin} ({{nR}\over{R + h}}}) - \mathrm{arcsin}({{{R}\over R + h}}) $$ where \(R\) = 6378 km is the radius of the earth. (b) Calculate \(\delta\) using \(n\) = 1.0003 and \(h\) = 20 km. How does this compare to the angular radius of the sun, which is about one quarter of a degree? (In actuality a light ray from the sun bends gradually, not abruptly, since the density and refractive index of the atmosphere change gradually with altitude.)

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?

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.

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 \(\theta_a = 36.2{^\circ}\), the ray refracted into the water makes an angle of 49.8\(^\circ\) with the normal to the interface. What is the smallest value of the incident angle \(\theta_a\) for which none of the ray refracts into the water?

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