The diffraction grating · 衍射光栅
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| diffraction grating/dɪˈfrækʃn ˈɡreɪtɪŋ/ | 衍射光栅 | yǎn shè guāng shān |
| slits/slɪts/ | 狭缝 | xiá fèng |
| coherent/kəʊˈhɪərənt/ | 相干 | xiāng gān |
| monochromatic/ˌmɒnəʊkrəʊˈmætɪk/ | 单色 | dān sè |
| maxima/ˈmæksɪmə/ | 极大 | jí dà |
| order/ˈɔːdə/ | 级次 | jí cì |
| normal/ˈnɔːml/ | 法线 | fǎ xiàn |
| spectrum/ˈspektrəm/ | 光谱 | guāng pǔ |
The rainbow on a CD
- Tilt a CD in the light and you see bright rainbow colours.
- The disc's fine tracks act as a diffraction grating 衍射光栅, splitting the light by wavelength.
- Gratings are how we measure wavelengths precisely.
光盘上的彩虹
- 把一张 CD 在光下倾斜,你会看到明亮的彩虹色。
- 光盘上细密的轨道充当 衍射光栅(diffraction grating),把光按波长分开。
- 光栅是我们精确测量波长的方法。
What a grating is
- A grating has many equally spaced slits 狭缝 — often hundreds per millimetre.
- Each slit is a coherent 相干 source, and they all interfere together.
A diffraction grating splits monochromatic 单色 light into sharp maxima on a screen
什么是光栅
- 光栅有 许多 等间距的狭缝——常常每毫米几百条。
- 每条狭缝都是一个相干波源,它们全部一起干涉。
衍射光栅把单色光分成屏上清晰的极大
Why the grating gives sharp maxima · 为什么光栅给出尖锐的极大
Two waves add when in phase and cancel when out of phase — change the phase and watch the resultant. A grating's many slits make the bright fringes razor-sharp. · 两列波同相时相加,反相时抵消——改变相位,看合成波。光栅的众多狭缝使亮纹变得极其尖锐。
A diffraction grating is made of: · 衍射光栅由以下哪种构成:
Hundreds or thousands of equally spaced slits, each acting as a coherent source. · 成百上千条等间距的狭缝,每一条都是一个相干波源。
The grating equation
- Bright maxima 极大 appear at angles given by $d\sin\theta = n\lambda$.
- $d$ = slit spacing, $n = 0, 1, 2, \ldots$ is the order 级次, $\theta$ measured from the normal 法线 (the straight-through direction).
光栅方程
- 亮的极大出现在由 $d\sin\theta = n\lambda$ 给出的角度上。
- $d$ = 缝间距,$n = 0, 1, 2, \ldots$ 是级次,$\theta$ 从 法线(normal)(直通方向)量起。
Light hits a grating of slit spacing $2.0\ \mu\text{m}$. The first-order ($n=1$) maximum is where $\sin\theta = 0.30$. Find the wavelength, in nm. · 光照到缝间距 $2.0\ \mu\text{m}$ 的光栅上。一级($n=1$)极大在 $\sin\theta = 0.30$ 处。求波长(用 nm)。
$\lambda = \dfrac{d\sin\theta}{n} = \dfrac{2.0 \times 10^{-6} \times 0.30}{1} = 6.0 \times 10^{-7}\ \text{m} = 600\ \text{nm}$. · $\lambda = \dfrac{d\sin\theta}{n} = \dfrac{2.0 \times 10^{-6} \times 0.30}{1} = 6.0 \times 10^{-7}\ \text{m} = 600\ \text{nm}$。
Sharper than two slits
- With many slits, every "wrong" direction is cancelled by lots of slits.
- So a grating gives much sharper maxima than a double slit.
Young's double-slit experiment — the single slit makes the two slits coherent sources
比双缝更锐利
- 有了许多狭缝,每个"错误"的方向都被大量狭缝抵消。
- 所以光栅给出的极大比双缝 锐利得多。
杨氏双缝实验——单缝使两条缝成为相干波源
A diffraction grating gives sharper maxima than a double slit. · 衍射光栅给出比双缝更尖锐的极大。
Yes — many slits cancel every "wrong" direction, leaving narrow, bright maxima. · 是的——许多狭缝抵消了每个“错误”方向,留下窄而亮的极大。
Slit spacing and highest order
- $N$ lines per mm → $d = \dfrac{1}{N}\ \text{mm}$.
- Since $\sin\theta \le 1$, the highest order is $n_{\text{max}} = \left\lfloor \dfrac{d}{\lambda} \right\rfloor$.
缝间距与最高级次
- 每毫米 $N$ 条线 → $d = \dfrac{1}{N}\ \text{mm}$。
- 由于 $\sin\theta \le 1$,最高级次为 $n_{\text{max}} = \left\lfloor \dfrac{d}{\lambda} \right\rfloor$。
A grating has $500$ lines per mm. What is the slit spacing $d$, in µm? · 一个光栅每毫米有 $500$ 条线。缝间距 $d$ 是多少(用 µm)?
$d = \dfrac{1}{500}\ \text{mm} = 0.002\ \text{mm} = 2.0\ \mu\text{m}$. · $d = \dfrac{1}{500}\ \text{mm} = 0.002\ \text{mm} = 2.0\ \mu\text{m}$。
For a grating and wavelength with $\dfrac{d}{\lambda} = 3.27$, the highest order seen is: · 对于 $\dfrac{d}{\lambda} = 3.27$ 的光栅和波长,能看到的最高级数是:
$n_{\text{max}} = \left\lfloor \dfrac{d}{\lambda} \right\rfloor = \lfloor 3.27 \rfloor = 3$. Order 4 would need $\sin\theta > 1$. · $n_{\text{max}} = \left\lfloor \dfrac{d}{\lambda} \right\rfloor = \lfloor 3.27 \rfloor = 3$。4 级需要 $\sin\theta > 1$。
Finding a wavelength
- Shine the light straight at the grating and measure the angle of a maximum.
- Then $\lambda = d\sin\theta$ (use $n = 1$ for the first order); average over orders to reduce error.
测量波长
- 让光垂直射向光栅,测量某个极大的 角度。
- 然后 $\lambda = d\sin\theta$(一级用 $n = 1$);对多个级次取平均以减小误差。
To find a wavelength with a grating, you measure the ____ of a maximum. · 用光栅求波长时,你测量一个极大的 ____。
Measure $\theta$ for a known order, then $\lambda = \dfrac{d\sin\theta}{n}$. · 测量已知级数的 $\theta$,然后 $\lambda = \dfrac{d\sin\theta}{n}$。
Worked example: a grating with 500 lines per mm
Light of wavelength $720\ \text{nm}$ falls normally on a grating with $500$ lines per mm. Find the angle between the two second-order maxima, the highest order visible, and the wavelength that would put its third-order maxima at the same angle.
- Spacing: $d = \dfrac{1}{500}\ \text{mm} = 2.0 \times 10^{-6}\ \text{m}$.
- Second order: $\sin\theta = \dfrac{2 \times 720 \times 10^{-9}}{2.0 \times 10^{-6}} = 0.72$, so $\theta = 46^\circ$. The two second-order maxima sit either side of the centre, so the angle between them is $2 \times 46^\circ = 92^\circ$.
- Highest order: $\dfrac{d}{\lambda} = \dfrac{2.0 \times 10^{-6}}{720 \times 10^{-9}} = 2.8$, so $n_{\text{max}} = 2$ — the third order would need $\sin\theta > 1$.
- Same angle, third order: $3\lambda = 2 \times 720\ \text{nm}$, so $\lambda = 480\ \text{nm}$ (blue).
- Check: at the same angle $d\sin\theta$ is fixed, so $n\lambda$ is fixed: a higher order needs a proportionally shorter wavelength.
例题:每毫米 500 条线的光栅
波长 $720\ \text{nm}$ 的光垂直射到每毫米 $500$ 条线的光栅上。求两个二级极大之间的夹角、能看到的最高级次,以及能使 三级 极大出现在同一角度的波长。
- 间距: $d = \dfrac{1}{500}\ \text{mm} = 2.0 \times 10^{-6}\ \text{m}$。
- 二级: $\sin\theta = \dfrac{2 \times 720 \times 10^{-9}}{2.0 \times 10^{-6}} = 0.72$,所以 $\theta = 46^\circ$。两个二级极大分别在中心两侧,所以它们 之间 的夹角是 $2 \times 46^\circ = 92^\circ$。
- 最高级次: $\dfrac{d}{\lambda} = \dfrac{2.0 \times 10^{-6}}{720 \times 10^{-9}} = 2.8$,所以 $n_{\text{max}} = 2$——三级需要 $\sin\theta > 1$。
- 同一角度,三级: $3\lambda = 2 \times 720\ \text{nm}$,所以 $\lambda = 480\ \text{nm}$(蓝光)。
- 检查: 在同一角度 $d\sin\theta$ 固定,所以 $n\lambda$ 固定:更高的级次需要按比例更短的波长。
Light of wavelength $600\ \text{nm}$ falls normally on a grating with $400$ lines per mm. What is the angle, in degrees, between the two first-order maxima? · 波长 $600\ \text{nm}$ 的光垂直射到每毫米 $400$ 条线的光栅上。两个一级极大之间的夹角是多少度?
$d = \dfrac{1}{400}\ \text{mm} = 2.5 \times 10^{-6}\ \text{m}$; $\sin\theta = \dfrac{600 \times 10^{-9}}{2.5 \times 10^{-6}} = 0.24$, so $\theta = 13.9^\circ$ and the angle between the two maxima is $2\theta = 27.8^\circ$. · $d = \dfrac{1}{400}\ \text{mm} = 2.5 \times 10^{-6}\ \text{m}$;$\sin\theta = \dfrac{600 \times 10^{-9}}{2.5 \times 10^{-6}} = 0.24$,所以 $\theta = 13.9^\circ$,两个极大之间的夹角为 $2\theta = 27.8^\circ$。
$\theta$ is measured from the normal, not from the surface of the grating, and "the angle between the two first-order maxima" means $2\theta$. Round $\dfrac{d}{\lambda}$ down for the highest order — a $\sin\theta$ above $1$ has no angle, so that order simply does not exist. And convert "lines per mm" to a spacing in metres before you start.
$\theta$ 从 法线 量起,而不是从光栅表面量起;"两个一级极大之间的夹角"指的是 $2\theta$。求最高级次时把 $\dfrac{d}{\lambda}$ 向下 取整——$\sin\theta$ 大于 $1$ 没有对应的角度,所以那一级根本不存在。另外,开始计算前先把"每毫米线数"换算成以 米 为单位的间距。
In the grating equation $d\sin\theta = n\lambda$, the angle $\theta$ is measured from the surface of the grating. · 在光栅方程 $d\sin\theta = n\lambda$ 中,角 $\theta$ 从光栅表面量起。
$\theta$ is measured from the normal — the straight-through direction of the zero order. Measuring from the surface swaps sine for cosine and gives the wrong wavelength. · $\theta$ 从法线——零级的直通方向——量起。从表面量会把正弦换成余弦,得出错误的波长。
White light through a grating
- The zero order is white: every wavelength has $\theta = 0$ there.
- Every other order is a spectrum 光谱, with violet nearest the centre (shortest $\lambda$, smallest $\theta$) and red furthest out.
- Higher orders are wider, and from the second order onwards they can overlap — red of order 2 can land on violet of order 3.
白光通过光栅
- 零级是 白色 的:每种波长在那里都有 $\theta = 0$。
- 其他每一级都是一个 光谱,紫色最靠近中心(最短的 $\lambda$,最小的 $\theta$),红色最靠外。
- 更高的级次更宽,从二级起可能 重叠——二级的红光可能落在三级的紫光上。
White light passes through a grating. Put these colours of the first-order spectrum in order, starting nearest the central maximum. · 白光通过光栅。把一级光谱中的这些颜色按顺序排列,从最靠近中央极大的开始。
From $d\sin\theta = n\lambda$, a shorter wavelength gives a smaller angle, so violet is nearest the centre and red furthest out. · 由 $d\sin\theta = n\lambda$,波长越短角度越小,所以紫色最靠近中心,红色最靠外。
You've got it
- a grating is many equally spaced slits → sharp maxima
- grating equation $d\sin\theta = n\lambda$, with $\theta$ from the normal; $d = \dfrac{1}{N}$ from "$N$ lines per mm"
- highest order $n_{\text{max}} = \left\lfloor \dfrac{d}{\lambda} \right\rfloor$; white light gives a spectrum in each order with violet nearest the centre
你掌握了
- 光栅是 许多 等间距的狭缝 → 锐利 的极大
- 光栅方程 $d\sin\theta = n\lambda$,$\theta$ 从 法线 量起;由"每毫米 $N$ 条线"得 $d = \dfrac{1}{N}$
- 最高级次 $n_{\text{max}} = \left\lfloor \dfrac{d}{\lambda} \right\rfloor$;白光在每一级都给出光谱,紫色最靠近中心