Refractometry is based on the difference in the speed of light through a substance (v) and through a vacuum ( \(c\) ). In the procedure, light of known wavelength passes through a fixed thickness of the substance at a known temperature. The index of refraction equals \(c / v\). Using yellow light \((\lambda=589 \mathrm{nm})\) at \(20^{\circ} \mathrm{C}\). for example, the index of refraction of water is 1.33 and that of diamond is \(2.42 .\) Calculate the speed of light in (a) water and (b) diamond.

Short Answer

Expert verified
a) ~2.26 x 10^8 m/s. b) ~1.24 x 10^8 m/s.

Step by step solution

01

- Understand the relationship

The index of refraction, denoted as n, is related to the speed of light in a vacuum (c) and the speed of light in the substance (v) by the equation: \[ n = \frac{c}{v} \].
02

- Determine the speed of light in vacuum

The speed of light in a vacuum is a known constant: \[ c = 3 \times 10^8 \, \text{m/s} \].
03

- Solve for the speed of light in water

Given the index of refraction of water, \[ n_{\text{water}} = 1.33 \]. Using the equation \( n = \frac{c}{v} \), solve for v: \[ v_{\text{water}} = \frac{c}{n_{\text{water}}} = \frac{3 \times 10^8 \, \text{m/s}}{1.33} \].
04

- Calculate the speed of light in water

Perform the division to find \[ v_{\text{water}} \approx 2.26 \times 10^8 \, \text{m/s} \].
05

- Solve for the speed of light in diamond

Given the index of refraction of diamond, \[ n_{\text{diamond}} = 2.42 \]. Using the same equation, solve for v: \[ v_{\text{diamond}} = \frac{c}{n_{\text{diamond}}} = \frac{3 \times 10^8 \, \text{m/s}}{2.42} \].
06

- Calculate the speed of light in diamond

Perform the division to find \[ v_{\text{diamond}} \approx 1.24 \times 10^8 \, \text{m/s} \].

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Key Concepts

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

Refractometry
Refractometry is a technique used to measure how light bends, or refracts, as it passes from one medium to another. This bending occurs because light changes speed when it moves between different substances. The amount of bending, or refraction, provides valuable information about the substance's properties.
In a refractometry experiment, light of a known wavelength travels through a substance at a controlled temperature. The main goal is to determine the speed of light inside the substance compared to its speed in a vacuum.
This method is particularly useful in identifying materials and assessing their purity. For instance, it is widely used in the food and beverage industry to measure sugar content in liquids.
Index of Refraction
The index of refraction, denoted as \( n \), measures how much light slows down when passing through a substance. It is calculated using the formula: \[ n = \frac{c}{v} \], where \( c \) is the speed of light in a vacuum and \( v \) is the speed of light in the substance.
A higher value of \( n \) indicates that light slows down more in that substance. For example, the index of refraction for water is 1.33, meaning light travels 1.33 times slower in water than in a vacuum. Similarly, the index for diamond is 2.42, indicating an even greater reduction in speed.
Understanding the index of refraction is critical in various fields. It helps in designing lenses and glasses, optimizing optical instruments, and even in understanding natural phenomena like mirages.
Speed of Light in Vacuum
The speed of light in a vacuum, denoted as \( c \), is a fundamental constant of nature. Its value is approximately \( 3 \times 10^8 \text{ m/s} \). This speed is the maximum at which all energy, matter, and information in the universe can travel.
The importance of this constant cannot be overstated. It forms the basis of many equations in physics, such as Einstein's famous equation \( E = mc^2 \). Knowing the speed of light in a vacuum allows scientists to measure and understand time, distance, and the nature of the universe with incredible precision.
In refractometry, \( c \) provides a reference point for determining how much light slows down in different substances, enabling accurate measurement of the index of refraction.
Light Wavelength
Wavelength is a key characteristic of light, referring to the distance between consecutive peaks of a light wave. It is often denoted by the Greek letter lambda \( \lambda \).
Wavelengths of visible light range from about 400 nanometers (violet) to 700 nanometers (red). Different wavelengths correspond to different colors perceived by our eyes.
In refractometry, the wavelength of the light used can affect the accuracy of the measurements. For instance, yellow light with a wavelength of 589 nanometers is often used because it provides reliable and consistent results.
The relationship between wavelength and speed of light is given by the equation \[ c = \lambda f \], where \( f \) is the frequency of the light. This relationship helps in understanding how light interacts with different materials and affects its propagation speed.

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Most popular questions from this chapter

In the course of developing his model, Bohr arrived at the following formula for the radius of the electron's orbit: \(r_{n}=\) \(n^{2} h^{2} \varepsilon_{0} / \pi m_{0} e^{2},\) where \(m_{c}\) is the electron's mass, \(e\) is its charge, and \(\varepsilon_{0}\) is a constant related to charge attraction in a vacuum. Given that \(m_{\mathrm{z}}=9.109 \times 10^{-31} \mathrm{~kg}, e=1.602 \times 10^{-19} \mathrm{C},\) and \(\varepsilon_{0}=8.854 \times 10^{-12} \mathrm{C}^{2} / \mathrm{J} \cdot \mathrm{m}\) calculate the following: (a) The radius of the first \((n=1)\) orbit in the \(\mathrm{H}\) atom (b) The radius of the tenth \((n=10)\) orbit in the \(\mathrm{H}\) atom

The photoelectric effect is illustrated in a plot of the kinetic energies of electrons ejected from the surface of potassium metal or silver metal at different frequencies of incident light. (a) Why don't the lines begin at the origin? (b) Why don't the lines begin at the same point? (c) From which metal will light of shorter wavelength eject an electron? (d) Why are the slopes equal?

For each of the following sublevels, give the \(n\) and \(l\) values and the number of orbitals: (a) \(6 g ;(b) 4 s ;(c) 3 d\)

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