# 有關精密量測的題目

1.雷射掃描測量儀中 ,假設旋轉八面鏡的旋轉速度為6000rpm, 透鏡之焦距為8cm 若計數器的頻率為80MHz ,請問可量測之精確度為何?

2.有一光波測距儀 ,計數器之震盪頻率為1GHz ,試問其量測之精度為何?

3.一外差干涉儀採用Michelson干涉儀架構, 假設波長為632.8nm, 頻差為2.5MHz ,有一移動面鏡 ,求其最大移動速度為何?

4.同上題 ,若相位解晰度為0.01度, 請問可解析的最小移動量為何?

5.同上題 ,若此雷射之同調長度為1m ,請問其量測之範圍為何?

6.Doppler Effect當遠離光源時, 對觀察者而言其頻率是下降或上升?假設測速照相機之調制頻率300MHz ,請問可測量之最大速度為何?

7.請設計一種可以測量輪廓的光學儀器

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Lv 4

1. f=80MHz , t=1/80MHz=12.5 (ns)

If let Δt=125 (ns) then to count "Δt/t" Numeric data .

Error Numeric data ( ±t count) → ±12.5 (ns)

rpm (revolution per minute) , given 6000 rpm

Conversion rpm to rps : 6000/60=100 rps

ω= 2π* rps = 2π* 100 = 628.318 (rad/s)

v=Rω= 0.08 * 2π* 100 = 50.265 (m/s)

Error : Δd = ± v * t = ( ± 50.265 m/s ) * ( 12.5 ns ) = ± 0.6283125 um

d = v * Δt = ( 50.265 m/s ) * ( 125 ns ) = 6.283125 um

Exactness : D = d ± Δd = 6.283125 ± 0.6283125 um

2. given frequency counter f=1GHz

t= 1/f = 1/1GHz = 1ns , L(min)= C*Δt / 2

L(min)= C*Δt / 2 = 0.3G (m/s) *1n (s) / 2 = 0.15 m = 15 cm .

3. by Δf = 2v / λ given Δf= 2.5MHz and λ= 632.8nm

then V(max) = Δf *λ/ 2 = 2.5MHz * 632.8nm / 2 = 0.791 (m/s)

4. If ΔΦ = 0.01 then Δd = ?

by ΔΦ= K * δ= ( 2π/λ ) * n * 2Δd = ( 4π/λ) Δd ( n=1.002 in air)

Δd= ΔΦ * λ / 4π = 0.01 * 632.8 nm / 4π = 0.503 nm

it is a micro displacement .

5. If coherence length Lc = 1m

ΔΦ= ( 2π/λ ) * n * 2Δd ≦ 2π

OPL ≦ Coherence length ( i.e. Δd ≦ λ/ 2 if in the air .)

Δd ≦ λ/ 2 → Δd ≦ Lc / 2 → Δd ≦ 1m / 2 → Δd ≦ 0.5m .

6. when Doppler Effect far away the source , frequency is decreased

wavelength is increased and the space is stretched .

Given frequency Δf =300MHz

by Δf = 2v / λ , V(max) = Δf *λ/ 2 = 300Mhz * λ/ 2 = ( 150MHz * λ ) m/s .

7. Use the Scanning techniques to design Isometric plot of Union of Jack method of flatness measurement .

2006-04-14 23:22:07 補充：

[correction] 4 .If ΔΦ = 0.01 then Δd = ?ΔΦ = 1/3600 *π/180 * 0.01 = 48.48n (rad)by ΔΦ= K * δ= ( 2π/λ ) * n * 2Δd = ( 4π/λ) Δd ( n=1.002 in air)Δd= ΔΦ * λ / 4π = 48.48n * 632.8 nm / 4π = 0.0024 pm it is a pico displacement .

2006-04-15 01:04:51 補充：

you're welcome !!! Good luck to your Midterm ...

2006-04-20 19:30:37 補充：

[correction] 4 .

If ΔΦ = 0.01 then Δd = ?

ΔΦ = 0.01 * π/180 = 174.53 u (rad)

by ΔΦ= K * δ= ( 2π/λ ) * n * 2Δd = ( 4π/λ) * Δd ( n=1.002 in air)

Δd= ΔΦ * λ / 4π = 174.53 u * 632.8 nm / 4π = 8.788 pm

it is a pico displacement .

2006-04-20 19:48:48 補充：

(A).

E1 = A cos( wt -Φ1 ) , E2 = B cos( wt-Φ2 )

I = Ia + Ib + ( 2√IaIb ) cos (Φ2 – Φ1)

= Io [ 1 + V cos ΔΦ ]

Let Io = Ia + Ib ， V= ( 2√IaIb ) / Ia + Ib ， ΔΦ = Φ2 – Φ1

“ ( 2√IaIb ) cos (Φ2 – Φ1) “ called Interference term.

2006-04-20 19:50:41 補充：

(B).

Equally ,

E1 = A cos ( w1t -Φ1 ) , E2 = B cos( w2t-Φ2 )

I = Io [ 1 + V cos ( Δwt + ΔΦ) ]

Let Io = Ia + Ib ， V= ( 2√IaIb ) / Ia + Ib ，Δw = w1 – w2 ， ΔΦ = Φ2 – Φ1

2006-04-20 19:51:15 補充：

“( 2√IaIb ) cos ( Δwt + ΔΦ) “ called Interference term.

If Δw = 0 then I = Io [ 1 + V cos ΔΦ ] same of (A).

Source(s): me

第一題掃瞄次數要乘八吧。。。八面鏡

這個我是第一次用雅虎知識問人　沒想到真的有好心人幫我回答　突然想起咖啡廣告的台詞you are so beautiful 真的很謝謝你

2006-04-15 00:45:50 補充：

不會啦 聽我修他實習的同學說他實習課都會在實習教室 不像其他老師都跑回去辦公室休息 現在這種老師很少了 這才是愛台灣拉

2006-04-15 13:11:51 補充：

不好意思　可以在請問你一題嗎

1.兩光之電場強度大小為E1=A( wt -Φ1 ),E2=B cos( wt-Φ2 ),求完全重疊後之干涉數學式 ？

2.同上題,若兩光有些微的頻差f,請問此數學式應為何？

2006-04-15 13:23:49 補充：

還有請問一下第一題得Δt=125 (ns)

這個是怎麼來的，是自己假設的嗎？