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

Problem 20b

Chapter 35, Problem 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?

Verified step by step guidance

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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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

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.

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Textbook Question

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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Textbook Question

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 first minimum

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Textbook Question

A uniform film of TiO₂, 1036 nm thick and having index of refraction 2.62, is spread uniformly over the surface of crown glass of refractive index 1.52. Light of wavelength 520.0 nm falls at normal incidence onto the film from air. You want to increase the thickness of this film so that the reflected light cancels. What is the minimum thickness of TiO₂ that you must add so the reflected light cancels as desired?

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Textbook Question

Coherent light of frequency 6.32 × 10¹⁴ Hz passes through two thin slits and falls on a screen 85.0 cm away. You observe that the third bright fringe occurs at ±3.11 cm on either side of the central bright fringe. (a) How far apart are the two slits? (b) At what distance from the central bright fringe will the third dark fringe occur?

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Textbook Question

When viewing a piece of art that is behind glass, one often is affected by the light that is reflected off the front of the glass (called glare), which can make it difficult to see the art clearly. One solution is to coat the outer surface of the glass with a film to cancel part of the glare. If the glass has a refractive index of 1.62 and you use TiO₂, which has an index of refraction of 2.62, as the coating, what is the minimum film thickness that will cancel light of wavelength 505 nm?

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Textbook Question

What is the thinnest film of a coating with n = 1.42 on glass (n = 1.52) for which destructive interference of the red component (650 nm) of an incident white light beam in air can take place by reflection?

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