课程： 5.2 波长和频率

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波长和频率
Do you enjoy going to the beach?
::你喜欢去海滩吗?

During the summer, almost everyone enjoys going to the beach. They can swim, have picnics, and work on their tans. But if you get too much sun, you can burn. A particular set of solar wavelengths are especially harmful to the skin. This portion of the solar spectrum is known as UV B, with wavelengths of 280-320 nm. Sunscreens are effective in protecting the skin against both immediate skin damage and the long-term possibility of skin cancer.
::夏天,几乎每个人都喜欢去海滩。他们可以游泳, 野餐, 晒太阳。但如果太阳太长, 你可以燃烧。一组特定的太阳波长对皮肤特别有害。这部分太阳光谱被称为紫外线B, 波长280-320纳米。太阳屏幕可以有效保护皮肤免受皮肤直接损伤和长期皮肤癌的可能性。

Waves
::波浪

Waves are characterized by their repetitive motion. Imagine a toy boat riding the waves in a wave pool. As the water wave passes under the boat, it moves up and down in a regular and repeated fashion. While the wave travels horizontally, the boat only travels vertically up and down. The Figure shows two examples of waves.
::海浪的特征是重复运动。想象一下, 一只玩具船在海浪中乘着海浪在海浪池中航行。当水浪从船底经过时, 水浪会以固定和反复的方式向上和向下移动。虽然海浪水平移动, 船只垂直向上和向下移动。图显示了两个海浪的例子。

(A) A wave consists of alternation crests and troughs. The wavelength (λ) is defined as the distance between any two consecutive identical points on the waveform. The amplitude is the height of the wave. (B) A wave with a short wavelength (top) has a high frequency because more waves pass a given point in a certain amount of time. A wave with a longer wavelength (bottom) has a lower frequency.

A wave cycle consists of one complete wave – starting at the zero point, going up to a wave crest , going back down to a wave trough , and back to the zero point again. The wavelength of a wave is the distance between any two corresponding points on adjacent waves. It is easiest to visualize the wavelength of a wave as the distance from one wave crest to the next. In an equation, wavelength is represented by the Greek letter lambda $\begin{align*}(\lambda)\end{align*}$ . Depending on the type of wave, wavelength can be measured in meters, centimeters, or nanometers (1 m = 10 ⁹ nm). The frequency , represented by the Greek letter nu $\begin{align*}( \nu )\end{align*}$ , is the number of waves that pass a certain point in a specified amount of time. Typically, frequency is measured in units of cycles per second or waves per second. One wave per second is also called a Hertz (Hz) and in SI units is a reciprocal second (s ^-1 ).
::一个波周期由一个完整的波组成——从零点开始,升到波顶,再回到波槽,再回到零点。波的波长是相邻波的两个相应点之间的距离。最容易将波的波长作为从一个波柱到下一个波峰的距离。在一个方程中,波长由希腊字母羊羔()代表。根据波的类型,波长可以以米、厘米或纳米计(1米=109纳米)测量。以希腊字母nu(Vu)表示的波的频率是在特定时间内通过某一点的波的频率。通常,波的频率以每秒或每秒的波的周期单位计量。每秒一个波也称为赫茨(Hz),在SI单位中,波长是对应的第二(s-1)。

Figure B above shows an important relationship between the wavelength and frequency of a wave. The top wave clearly has a shorter wavelength than the second wave. However, if you picture yourself at a stationary point watching these waves pass by, more waves of the first kind would pass by in a given amount of time. Thus the frequency of the first waves is greater than that of the second waves. Wavelength and frequency are therefore inversely related. As the wavelength of a wave increases, its frequency decreases. The equation that relates the two is:
::以上图B显示了波浪波长和波频率之间的重要关系。顶波的波长显然比第二波短。但是, 如果你在固定的点看到这些波经过, 就会在一定的时间内通过更多的第一种波。因此, 第一波的频率大于第二波的频率。因此波长和频率是反向相关的。随着波的波长增加,波的频率会下降。与这两个波相关的方程式是:

$\begin{aligned} c = λ ν \end{aligned}$
$\begin{align*}\ c = \lambda \nu\end{align*}$
::

The variable $\begin{align*}c\end{align*}$ is the speed of light . For the relationship to hold mathematically, if the speed of light is used in m/s, the wavelength must be in meters and the frequency in Hertz.
::变量c 是光速。如果光速用于 m/ s 中, 则波长必须以米和 Hertz 的频率为单位。

Sample Problem: Wavelength and Frequency
::问题:波长和频率

The color orange within the visible light spectrum has a wavelength of about 620 nm. What is the frequency of orange light?
::可见光谱中的橙色的波长约为620纳米。橙色光的频率是多少?

Step 1: List the known quantities and plan the problem.
::第1步:列出已知数量并规划问题。

Known
::已知已知

wavelength $\begin{align*}(\lambda)\end{align*}$ = 620 nm
:) = 620纳米

speed of light $\begin{align*}(c)\end{align*}$ = 3.00 × 10 ⁸ m/s
:c) = 3.00 × 108米/秒

conversion factor 1 m = 10 ⁹ nm
::换算系数 1 m = 109 nm

Unknown
::未知

Frequency
::频率频率频率

Convert the wavelength to m, then apply the equation $\begin{align*}c=\lambda\nu\end{align*}$ and solve for frequency. Dividing both sides of the equation by $\begin{align*}\lambda\end{align*}$ yields:
::将波长转换为 m, 然后应用方程式 c 和频率解析。将方程式两边区分为产值 :

$\begin{aligned} ν = \frac{c}{λ} \end{aligned}$
$\begin{align*}\nu=\frac{c}{\lambda}\end{align*}$
::~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Step 2: Calculate.
::第2步:计算。

$\begin{aligned} 620 nm \times (\frac{1 m}{10^{9} nm}) & = 6.20 \times 10^{- 7} m \\ ν = \frac{c}{λ} = \frac{3.0 \times 10^{8} m/s}{6.20 \times 10^{- 7} m} & = 4.8 \times 10^{14} Hz \end{aligned}$
$\begin{align*}620 \ \text{nm} \times \left(\frac{1 \ \text{m}}{10^9 \ \text{nm}}\right) &=6.20 \times 10^{-7} \ \text{m}\\ \nu = \frac{c}{\lambda}=\frac{3.0 \times 10^8 \ \text{m/s}}{6.20 \times 10^{-7} \ \text{m}} &=4.8 \times 10^{14} \ \text{Hz}\end{align*}$
::620 nmxx( 1 m109 nm) = 6.20×10- 7 mc3. 0×108 ms 6.20×10- 7 ms= 4.8×1014 Hz

Step 3: Think about your result.
::步骤3:想想你的结果。

The value for the frequency falls within the range for visible light.
::频率值在可见光范围以内。

Summary
::摘要

All waves can be defined in terms of their frequency and intensity.
::所有波浪都可以根据其频率和强度来界定。

$\begin{align*}c=\lambda\nu\end{align*}$ expresses the relationship between wavelength and frequency.
::c 表示波长和频率之间的关系。

Review
::回顾

Define wavelength.
::定义波长。

Define frequency.
::界定频率。

What is the relationship between wavelength and frequency?
::波长和频率之间的关系是什么?

章节大纲

常规

波长和频率

Do you enjoy going to the beach? ::你喜欢去海滩吗?

Waves ::波浪

Sample Problem: Wavelength and Frequency ::问题:波长和频率

Summary ::摘要

Review ::回顾

Do you enjoy going to the beach?
::你喜欢去海滩吗?

Waves
::波浪

Sample Problem: Wavelength and Frequency
::问题:波长和频率

Summary
::摘要

Review
::回顾