Съдържание

USAPhO 1999, Полуфинал — Задача 3. A3 · Young’s Triple Slit

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Всички материали от USAPhO 1999

1999 Semi-Final Exam

Условие

Three slits separated by d and d on a barrier at the left; three rays from the slits converge at point P on a screen at distance D on the right; the angle θ is marked between the middle ray and the horizontal.

Coherent light of wavelength $\lambda$ passes through three infinitesimally narrow slits with spacing $d$ . It produces an interference pattern on screen a large distance $D \gg d$ away. (The figure above has been distorted to fit the page.) Assume the light through all three slits is in phase at the location of the slits. Further assume that a wave through any slit has amplitude $A$ at any point $P$ on the screen.

a. What is the phase shift between adjacent waves arriving at point $P$ as a function of θ? [3 т.]

b. Write a wave function for each of the three waves arriving at point P. Specify which wave is which by using the subscripts “t”, “m”, and “b” to denote the wave through the top, middle, and bottom slit, respectively. Define any quantities you introduce. [3 т.]

c. What is the amplitude of the resultant wave at point P? [7 т.]

d. Write an expression for the intensity on the screen as a function of angle θ. Let $I_o$ be the intensity at θ = 0. [3 т.]

e. What is the position of the first minimum, that is, what is the smallest angle at which the intensity is zero? [4 т.]

(contributed by Mary Mogge)

Отговори

Решение

a. The path difference $\Delta L$ between wave through adjacent slits is

\Delta L = d\sin\theta.

The phase difference δ is

\delta = \frac{2\pi}{\lambda}\Delta L = \frac{2\pi}{\lambda}d\sin\theta

\Psi_t = A\sin\left(\omega t + \phi - \delta\right)

where ω = 2πf, f is the frequency of the wave and ϕ is a constant that is independent of the wave (may be set to zero).

\Psi_m = A\sin\left(\omega t + \phi\right)

\Psi_b = A\sin\left(\omega t + \phi + \delta\right)

(or any equivalent set with phase shifts differing by δ).

\Psi = \Psi_t + \Psi_m + \Psi_b

\Psi = A\sin\left(\left(\omega t + \phi\right) - \delta\right) + A\sin\left(\omega t + \phi\right) + A\sin\left(\left(\omega t + \phi\right) + \delta\right)

\Psi = A\left[\sin\left(\omega t + \phi\right)\cos\delta - \sin\delta\cos\left(\omega t + \phi\right)\right] + A\sin\left(\omega t + \phi\right)

+ A\left[\sin\left(\omega t + \phi\right)\cos\delta + \sin\delta\cos\left(\omega t + \phi\right)\right]

\Psi = A\left(1 + 2\cos\delta\right)\sin\left(\omega t + \phi\right) = A_{res}\sin\left(\omega t + \phi\right)

With the resultant amplitude

A_{res} = A\left(1 + 2\cos\delta\right)

d. The intensity is proportional to the amplitude squared. Calling the constant of proportionality c, we have

I(\theta) = cA^2\left(1 + 2\cos\delta\right)^2.

At $\theta = 0$ , $\delta = 0$ , $\cos\delta = 1$

I_o = cA^2\left(1 + 2\right)^2 = 9cA^2.

Eliminating c from the $I(\theta)$ equation

I(\theta) = \frac{I_o}{9}\left(1 + 2\cos\delta\right)^2.

e. In order for $I(\theta) = 0$

1 + 2\cos\delta = 0

\cos\delta = -\tfrac{1}{2} \qquad \text{or} \qquad \delta = 120^{\circ} = 2\pi/3

\frac{2\pi}{3} = \frac{2\pi}{\lambda}d\sin\theta

\theta = \sin^{-1}\left(\frac{\lambda}{3d}\right)

Two adjacent slits a distance d apart with parallel rays leaving at angle θ; a perpendicular dropped from the upper ray marks the path difference ΔL on the lower ray, with the angle θ shown at the slit.

Автори

The 1999 Semi-Final Examination questions were written by the coaches of the United States Physics Team. The coaches are: Academic Director, Dr. Mary Mogge – Professor of Physics at California State Polytechnic University, Pomona, CA; Senior Coach, Dr. Leaf Turner – Physicist in the Theoretical Division of Los Alamos National Laboratory, Los Alamos, NM; Dr. Warren Turner – Physics Teacher at St. Paul's School, Concord, NH. The coaches would also like to thank former academic director, Dr. Larry Kirkpatrick for his many helpful comments.

Оригинал в Архива: Semifinal 1999.pdf · официални решения: Semifinal 1999 Solutions.pdf