Ch. 34 - The Wave Nature of Light: Interference and Polarization

Giancoli Douglas - Physics for Scientists and Engineers 5th edition

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Giancoli Douglas 5th edition

Ch. 34 - The Wave Nature of Light: Interference and Polarization

Problem 69

Chapter 33, Problem 69

Light of wavelength 690 nm passes through two narrow slits 0.66 mm apart. The screen is 1.75 m away. A second source of unknown wavelength produces its second-order fringe 1.23 mm closer to the central maximum than the 690-nm light. What is the wavelength of the unknown light?

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Identify the given values for the first light source: wavelength \\(\lambda_1 = 690 \\, \text{nm}\\), slit separation \\(\text{d} = 0.66 \\, \text{mm}\\), and distance to the screen \\(\text{L} = 1.75 \\, \text{m}\\).

Use the formula for the position of the m-th order bright fringe for a double-slit interference pattern: \\(\text{y}_m = \frac{m \lambda L}{d}\\), where \\(\text{y}_m\\) is the fringe position, \\(\text{m}\\) is the order of the fringe, \\(\lambda\\) is the wavelength, \\(\text{L}\\) is the distance to the screen, and \\(\text{d}\\) is the slit separation.

Calculate the position of the second-order (\\(\text{m} = 2\\)) bright fringe for the 690 nm light using the above formula.

For the second light source, note that its second-order fringe is 1.23 mm closer to the central maximum than that of the 690 nm light. Let \\(\lambda_2\\) be the wavelength of the unknown light. Set up the equation \\(\frac{2 \lambda_2 L}{d} = \frac{2 \cdot 690 \\, \text{nm} \cdot L}{d} - 1.23 \\, \text{mm}\\) to find \\(\lambda_2\\).

Solve the equation for \\(\lambda_2\\) to find the wavelength of the unknown light.

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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 light passes through two closely spaced slits. The resulting pattern consists of alternating bright and dark fringes on a screen, which can be analyzed to determine the wavelength of the light used. The position of these fringes is influenced by the distance between the slits, the distance to the screen, and the wavelength of the light.

Interference and Fringe Spacing

Interference occurs when two or more waves overlap, resulting in a new wave pattern. In the context of the double-slit experiment, the fringe spacing can be calculated using the formula: y = (m * λ * L) / d, where y is the fringe position, m is the order of the fringe, λ is the wavelength, L is the distance to the screen, and d is the distance between the slits. This relationship allows for the determination of unknown wavelengths based on the positions of the fringes.

Wavelength Calculation

Wavelength is a fundamental property of waves, defined as the distance between successive peaks of a wave. In this problem, the difference in fringe positions for two light sources allows for the calculation of the unknown wavelength using the known wavelength and the geometry of the setup. By applying the interference conditions and the relationship between fringe positions and wavelengths, one can derive the wavelength of the unknown light source.

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