Two slits spaced 0.260 mm apart are 0.900 m from a screen and illuminated by coherent light of wavelength 660 nm. The intensity at the center of the central maximum (u = 0°) is I0. What is the distance on the screen from the center of the central maximum to the point where the intensity has fallen to I0/2?

Ch 35: Interference

Young & Freedman Calc - University Physics 14th Edition

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Young & Freedman Calc 14th Edition

Ch 35: Interference

문제 20b

35장, 문제 20b

Two slits spaced 0.260 mm apart are 0.900 m from a screen and illuminated by coherent light of wavelength 660 nm. The intensity at the center of the central maximum (u = 0°) is I₀. What is the distance on the screen from the center of the central maximum to the point where the intensity has fallen to I₀/2?

검증된 단계별 안내

Step 1: Recognize that this is a double-slit interference problem. The intensity at a point on the screen is given by the formula: \( I = I_0 \cos^2(\phi/2) \), where \( \phi \) is the phase difference between the light waves from the two slits. The problem asks for the distance on the screen where the intensity falls to \( I_0/2 \).

Step 2: Set \( I = I_0/2 \) in the intensity formula. Solving \( \cos^2(\phi/2) = 1/2 \) gives \( \phi/2 = \pi/4 \) or \( \phi = \pi/2 \). This means the phase difference \( \phi \) at the desired point is \( \pi/2 \).

Step 3: The phase difference \( \phi \) is related to the path difference \( \Delta x \) by the formula \( \phi = 2\pi \Delta x / \lambda \), where \( \lambda \) is the wavelength of the light. Substituting \( \phi = \pi/2 \), solve for \( \Delta x \): \( \Delta x = \lambda/4 \).

Step 4: The path difference \( \Delta x \) is related to the position \( y \) on the screen by the geometry of the setup: \( \Delta x = d \sin(\theta) \), where \( d \) is the slit separation and \( \theta \) is the angle to the point on the screen. For small angles, \( \sin(\theta) \approx \tan(\theta) = y/L \), where \( L \) is the distance to the screen. Substituting \( \Delta x = \lambda/4 \), solve for \( y \): \( y = (\lambda L) / (4d) \).

Step 5: Substitute the given values into the formula \( y = (\lambda L) / (4d) \): \( \lambda = 660 \times 10^{-9} \) m, \( L = 0.900 \) m, and \( d = 0.260 \times 10^{-3} \) m. Simplify the expression to find the distance \( y \) on the screen where the intensity falls to \( I_0/2 \).

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Double-Slit Experiment

The double-slit experiment demonstrates the wave nature of light through interference patterns created when coherent light passes through two closely spaced slits. The resulting pattern consists of alternating bright and dark fringes on a screen, where the bright fringes correspond to constructive interference and the dark fringes to destructive interference. Understanding this experiment is crucial for analyzing how light behaves in the given scenario.

Interference and Intensity

Interference occurs when two or more waves overlap, leading to a new wave pattern. The intensity of light at any point on the screen is determined by the superposition of the light waves from the two slits. The intensity at a point can be expressed in terms of the maximum intensity and the angle of the point relative to the central maximum, which is essential for calculating the distance to the point where the intensity falls to half its maximum value.

Path Difference and Angular Position

The path difference between the light waves from the two slits is critical in determining the interference pattern. For small angles, the path difference can be approximated as the product of the slit separation and the sine of the angle from the central maximum. This relationship allows us to calculate the angular position of points on the screen where specific intensity values occur, such as where the intensity is half of the maximum.