Optical instruments, path differences and refraction
GRE Subject Test · GRE Physics Topic 17 5:26 English narration · English + 中文 subtitles burned in
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A gas cell in one arm of a Michelson interferometer is crossed twice, so counting only one passage halves the inferred change in refractive index.
迈克耳孙干涉仪一臂中的气室被光穿过两次，所以只计一次会使推得的折射率变化减半。
This topic computes telescope magnification and grating resolving power, relates fringe counts to double-pass path changes, and traces surface normals before applying Snell's law.
本主题计算望远镜放大率和光栅分辨本领，把条纹数与双程光程变化联系起来，并在应用斯涅耳定律之前先画出表面法线。
For an ideal astronomical telescope adjusted for relaxed viewing at infinity, the objective forms its focal plane image and the eyepiece collimates the emerging light.
对于针对无穷远放松观察调节的理想天文望远镜，物镜形成其焦平面像，目镜使出射光准直。
Lens separation is f sub objective plus f sub eyepiece.
透镜间距为f下标物镜加f下标目镜。
Angular magnification magnitude is f sub objective divided by f sub eyepiece.
角放大率大小为f下标物镜除以f下标目镜。
The usual two converging lens telescope produces an inverted image, represented by a negative signed angular magnification under a consistent convention.
通常的双凸透镜望远镜产生倒立像，在一致约定下用带负号的角放大率表示。
A finite final image distance changes lens separation.
有限最终像距会改变透镜间距。
Do not use the infinity adjustment without checking the condition.
在未检查条件之前，不要使用无穷远调节。
Resolving power R, lambda over delta lambda, describes the smallest distinguishable wavelength separation near lambda.
分辨本领 R 等于 lambda 除以 delta lambda，描述了在波长 lambda 附近可分辨的最小波长间隔。
For an ideal diffraction grating in order m with N illuminated slits, R equals m times N.
对于级次为 m、有 N 个被照亮狭缝的理想衍射光栅，R 等于 m 乘以 N。
This is a resolution criterion, not simply the angular position formula d sine theta equals m lambda.
这是一个分辨判据，而不仅仅是角位置公式 d sin theta 等于 m lambda。
Increasing illuminated slit count sharpens the principal peaks; increasing slit spacing mainly changes their angular locations.
增加被照亮的狭缝数量会使主极大峰变锐；增加狭缝间距主要改变它们的角位置。
A measured lambda of 600 nanometres and delta lambda of 3 nanometres gives R equal to 200, independent of converting both lengths to metres because their ratio uses matching units.
测得 lambda 为 600 纳米，delta lambda 为 3 纳米，则 R 为 200，这与是否将两个长度都转换为米无关，因为它们的比值使用匹配单位即可。
Michelson interference depends on the optical path difference between the arms.
迈克尔逊干涉取决于两臂之间的光程差。
If one arm contains a gas cell of geometric length L and index changes from n to 1 upon evacuation, a double passage changes path by 2L(n−1).
若其中一臂包含几何长度为 L 的气室，且抽真空后折射率从 n 变为 1，则双程通过会导致光程改变 2L(n−1)。
If N fringes pass, the magnitude of this change is Nλ, so n−1=Nλ/(2L).
若有 N 条条纹移过，该变化的量级为 Nλ，因此 n−1 = Nλ/(2L)。
For a moving mirror instead, displacement Δx gives path change 2Δx.
对于移动反射镜而言，位移 Δx 对应的光程变化为 2Δx。
Fringes count phase cycles; one fringe is one wavelength of optical path, not one wavelength of mirror motion.
条纹计数的是相位周期；一条条纹对应一个波长的光程，而非一个波长的镜面运动距离。
Snell’s law, n 1 sine theta 1 equals n 2 sine theta 2, uses angles from the local surface normal.
斯涅尔定律，n1 sin θ1 等于 n2 sin θ2，使用相对于局部表面法线的角度。
Reflection has equal incoming and outgoing angles from that same normal.
反射时，入射角与反射角相对于同一法线相等。
In a rectangular block, normals of neighbouring faces are perpendicular; an incidence angle measured from one normal is complementary to the angle of that same ray from the other.
在矩形块中，相邻面的法线互相垂直：从一条法线量得的入射角，与同一光线相对另一条法线的角度互为余角。
Draw the ray and normals before substitution.
代入数值前先画出光线和法线。
For incidence from index n into air, total internal reflection occurs only when sine theta is greater than 1 over n; equality is the critical grazing case.
光从折射率为 n 的介质射向空气时，只有当 sin θ 大于 n 分之一时才发生全内反射；相等时是临界掠射情况。
A stated partly refracted ray must satisfy the transmission condition.
题目说明有部分折射光线时，该光线必须满足透射条件。
A gas cell 0.10 metres long is crossed twice.
一个长 0.10 米的气室被光穿过两次。
If 120 fringes pass at 500 nanometres while it is evacuated, n minus 1 is 120 times 500 times 10 to the minus 9, over 2 times 0.10: 3.0 times 10 to the minus 4.
抽空过程中在 500 纳米波长下移过 120 条条纹，则 n 减 1 等于 120 乘 500 乘 10 的负 9 次方，再除以 2 乘 0.10：3.0 乘 10 的负 4 次方。
So n is 1.00030.
所以 n 等于 1.00030。
A grating with spacing d equals 2.0 micrometres illuminates N equals 5000 slits at wavelength 600 nm.
光栅间距 d 等于 2.0 微米，以波长 600 nm 照射 N 等于 5000 条狭缝。
Find the largest physically allowed order and its ideal resolving limit in wavelength.
求最大物理允许级次及其理想波长分辨极限。
State the different roles of spacing and illuminated slit count.
说明间距与受照狭缝数的不同作用。
The absolute value of m lambda divided by d is less than or equal to 1, which gives m less than or equal to 3.33, hence the largest integer order is 3.
|mλ/d| ≤ 1，得出 m ≤ 3.33，因此最大整数级次为 3。
R equals m times N equals 15000, and delta lambda equals lambda divided by R, which is 600 nanometres divided by 15000, equaling 0.040 nanometre.
R = mN = 15000，且 Δλ = λ/R = 600 nm / 15000 = 0.040 nm。
Spacing sets the order angles; illuminated slit count narrows the peaks and improves resolution.
间距决定级次角；被照亮的狭缝数量使峰变窄并提高分辨率。
Award order condition, order, resolving limit and distinction.
评分点包括：级次条件、级次、分辨极限及区别。
The ideal limit assumes uniform illumination.
理想极限假设均匀照明。
Use optical path, not just geometric distance.
使用光程，而不仅仅是几何距离。
Refraction angles are measured from the normal of the actual surface, and telescope focal-length addition assumes a final image at infinity.
折射角是从实际表面的法线测量的，望远镜焦距相加假设最终像在无穷远处。
resolving power: Wavelength divided by the smallest resolvable wavelength separation.
分辨本领。 波长除以最小可分辨的波长间隔。
optical path length: Geometric path weighted by refractive index along the ray.
光程。 沿光线方向，几何路径乘以折射率。
Retrieve the meanings without reading the definitions, then change a parameter in the original transfer and test the limiting case.
在不看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。