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Ch 35: Interference
Young & Freedman Calc - University Physics 15th Edition
Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoYoung & Freedman Calc 15th EditionCh 35: InterferenceProblema 18b
Capitolo 34, Problema 18b

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?

Guida verificata passo dopo passo
1
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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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.
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Young's Double Slit Experiment

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.
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Wave Intensity

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.
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Mean Free Path
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In a two-slit interference pattern, the intensity at the peak of the central maximum is I0. At a point in the pattern where the phase difference between the waves from the two slits is 60.0°, what is the intensity?

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