hgbn asked in 科學其他:科學 · 6 years ago

急需高手,三題英文物理問題(20點)

1. Using monochromatic light of 410 nm wavelength, a single thin slit forms a diffraction pattern, with the first minimum at an angle of 40.0 from central maximum. The same slit, when illuminated by a new wavelength, produces a diffraction pattern with the second minimum at 60 angle from the central maximum. What is the wavelength if this new light?

A. 342 nm

B. 309 nm

C. 326 nm

D. 276 nm

E. 293 nm

2. A light beam with wavelength 500 nm us reflected constructively from a thin layer of oil having index of refraction 1.25. The oil floats on the top of water of index of refraction 1.33. What is the minimum thickness of the layer of oil?

A. 100 nm

B. 200 nm

C. 225 nm

D. 150 nm

E. 175 nm

3. A particle is moving at 0.86c. By what magnitude percentage is the Newtonian expression (p=m0v) for momentum in error? (The percentage error is the difference between the erroneous and correct expressions, relative the correct one).

A. 49%

B. 62%

C. 40%

D. 55%

感激不盡

Update:

謝謝!

還有一題需要幫忙:

圖在這網站裡:

http://www.chegg.com/homework-help/questions-and-a...

Update 2:

題目:

As shown in the figure, the orientation of the transmission axis for each of three ideal polarizing sheets is labeled relative to the vertical direction. A beam of light, polarized in the vertical direction, is incident on the first polarizer

with an intensity of 1.00 kW/m2.

Update 3:

What is the intensity of the beam after it has passed through the three polarizing sheets when θ1 = 30°, θ2 = 30°, and θ3= 60°?

A) 563 W/m2

B) 141 W/m2

C) 316 W/m2

D) 188 W/m2

E) 433 W/m2

感激不盡

Update 4:

還有可以再幫我這一題嗎:

http://tw.knowledge.yahoo.com/question/question?qi...

謝謝您

非常謝謝

1 Answer

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    Lv 7
    6 years ago
    Best Answer

    這些應該是普物的習題Q1:單狹縫實驗,暗紋公式:b*sinθ = nλ (b:狹縫寬度;λ:波長;n正整數,第n條暗紋)波長410 nm,第1條暗紋40度b*sin40 = 1*410波長λ’ nm,第2條暗紋60度b*sin60 = 2*λ’兩式相除得λ’ = 276.2 (nm) → D Q2:薄膜干涉,疏→密→更密,建設性反射:2t = m(λ/n) (λ:波長;n:折射率;m正整數,第m條暗紋)2t = 1 * (500/1.25)t = 200 (nm) → B Q3:相對論動量P = m0*V/√[ 1 – (V/C)^2]= m0*V/√[ 1 –0.86^2] = 1.96 m0V牛頓動量P = m0*Ve = (1.96 m0V - m0V) / 1.96 m0V = 0.4897 → A

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