Ondas y óptica
GRE Subject Test · GRE Physics Tema 3 3:48 Narración en inglés · Subtítulos en inglés + 中文 quemados en pantalla
Capítulos
Transcripción
Two individually bright waves can cancel at a point if their phases are opposite: crest meets trough.
两列各自明亮的波，如果在某点相位相反，就会相互抵消：波峰遇上波谷。
This topic covers superposition and interference; diffraction, polarisation and optical imaging; and how phase, frequency and dispersion are related.
本主题涵盖波的叠加与干涉；衍射、偏振和光学成像；以及相位、频率与色散之间的关系。
Add wave amplitudes, then calculate intensity.
先叠加波振幅，再计算强度。
Intensity is proportional to squared amplitude for the same medium.
在同一介质中，强度与振幅的平方成正比。
A path difference delta gives a phase difference of 2 pi delta over lambda.
路径差会产生相位差，即二派德尔塔除以拉姆达。
Coherent sources need a stable relative phase.
相干光源需要稳定的相对相位。
For a single slit, minima satisfy a sin θ=mλ for nonzero integers m.
对于单缝，极小值满足 a sin θ = mλ，其中 m 为非零整数。
A double slit has a different interference condition.
双缝具有不同的干涉条件。
Geometric optics needs a stated sign convention.
几何光学需要明确的符号约定。
Wave effects matter when apertures are comparable to wavelength.
当孔径与波长相当时，波动效应才显著。
Coherent waves have a stable phase difference.
相干波具有稳定的相位差。
Add amplitudes before squaring.
先叠加振幅，再进行平方运算。
For two coherent waves the intensities do not simply add.
两列相干波的光强并不简单相加。
The total is I one plus I two, plus a cross term: two root I one I two, times cos delta, where delta is the phase difference.
总光强等于 I₁ 加 I₂，再加一个交叉项：2 乘 I₁I₂ 的平方根再乘 cos δ，其中 δ 是相位差。
Take equal intensities I zero and a path difference of a quarter wavelength.
取两列光强都等于 I₀，光程差为四分之一波长。
The phase difference is two pi times delta over lambda: two pi times a quarter, which is pi over two.
相位差等于 2π 乘以光程差除以波长，即 2π 乘四分之一，等于 π/2。
With equal intensities the total is two I zero times one plus cos delta.
两列光强相等时，总光强等于 2I₀ 乘以 1 加 cos δ。
At pi over two the cosine is zero, so the intensity is two I zero: halfway between the dark minimum of zero and the bright maximum of four I zero.
在 π/2 处余弦为零，所以光强为 2I₀，正好介于暗纹最小值零和亮纹最大值 4I₀ 之间。
This is neither a maximum nor a minimum.
这既不是极大值，也不是极小值。
Random relative phase removes the cross term in a long time average.
随机相对相位会使长时间平均中的交叉项消失。
Stable coherence is a physical condition, not a drawing convention.
稳定的相干性是一种物理条件，而非绘图惯例。
Two coherent beams have individual intensities I and 4I in the same polarisation.
两束相干光束具有相同的偏振，其单独强度分别为I和4I。
Find the largest and smallest combined intensity, and the fringe visibility, I sub max minus I sub min, all over I sub max plus I sub min.
求最大和最小合成强度，以及条纹可见度，即I下标max减I下标min，除以I下标max加I下标min。
Explain why a dark fringe is not perfectly dark.
解释为什么暗条纹不是完全黑暗的。
Four marks.
四分。
Amplitudes scale as the square root of I, and two times the square root of I.
振幅与根号下I成正比，以及两倍根号下I成正比。
The maximum intensity equals the square root of I plus two times the square root of I, all squared, which equals nine I.
最大光强等于根号下I加两倍根号下I的平方，即9I。
The minimum intensity equals two times the square root of I minus the square root of I, all squared, which equals I.
最小光强等于两倍根号下I减根号下I的平方，即I。
Visibility equals nine I minus I, over nine I plus I, which equals zero point eight.
可见度等于9I减I除以9I加I，结果为0.8。
Unequal amplitudes cannot cancel completely.
振幅不等时无法完全抵消。
Award one mark each for maximum, minimum, visibility, and explanation.
对最大值、最小值、可见度及解释各给一分。
The central single-slit maximum is not obtained by setting m equal to zero in the minima condition.
中央单缝极大值并非通过将m设为零代入极小值条件而得到。
coherence: A stable phase relationship between waves.
相干性：波之间稳定的相位关系。
diffraction: Wave spreading and interference caused by an aperture or obstacle.
衍射：由孔径或障碍物引起的波扩散和干涉现象。
Retrieve the meanings without reading the definitions, then change a parameter in the original transfer and test the limiting case.
在不阅读定义的情况下回忆含义，然后改变原始传递中的一个参数并检验极限情况。