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 67

Chapter 33, Problem 67

Light of wavelength 5.0 x 10⁻⁷ passes through two parallel slits and falls on a screen 5.0 m away. Adjacent bright bands of the interference pattern are 2.0 cm apart.
(a) Find the distance between the slits.
(b) The same two slits are next illuminated by light of a different wavelength, and the fifth minimum for this light occurs at the same point on the screen as the fourth minimum for the previous light. What is the wavelength of the second source of light?

Verified step by step guidance

Step 1: For part (a), use the formula for the fringe spacing in a double-slit interference pattern: Δy = (λL) / d, where Δy is the fringe spacing (2.0 cm = 0.02 m), λ is the wavelength of the light (5.0 x 10⁻⁷ m), L is the distance to the screen (5.0 m), and d is the distance between the slits. Rearrange the formula to solve for d: d = (λL) / Δy.

Step 2: Substitute the given values into the formula for d: λ = 5.0 x 10⁻⁷ m, L = 5.0 m, and Δy = 0.02 m. Perform the calculation to find the distance between the slits, d.

Step 3: For part (b), recognize that the condition for minima in a double-slit interference pattern is given by the equation: d sin(θ) = mλ, where m is the order of the minimum, λ is the wavelength, and θ is the angle of the minimum. For the first light source, the fourth minimum corresponds to m₁ = 4, and for the second light source, the fifth minimum corresponds to m₂ = 5.

Step 4: Since the fifth minimum of the second light source coincides with the fourth minimum of the first light source, the path difference for both cases must be the same. Set up the equality: m₁λ₁ = m₂λ₂, where λ₁ is the wavelength of the first light (5.0 x 10⁻⁷ m) and λ₂ is the wavelength of the second light. Rearrange the equation to solve for λ₂: λ₂ = (m₁λ₁) / m₂.

Step 5: Substitute the known values into the equation for λ₂: m₁ = 4, λ₁ = 5.0 x 10⁻⁷ m, and m₂ = 5. Perform the calculation to find the wavelength of the second source of light, λ₂.

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Key Concepts

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

Young's Double Slit Experiment

Young's Double Slit Experiment demonstrates the wave nature of light through the interference pattern created when light passes through two closely spaced slits. The pattern consists of alternating bright and dark bands on a screen, resulting from constructive and destructive interference of light waves. The distance between the slits, the wavelength of the light, and the distance to the screen are crucial for calculating the positions of these bands.

Interference Pattern

An interference pattern is formed when two or more coherent light waves overlap, leading to regions of constructive interference (bright bands) and destructive interference (dark bands). The spacing of these bands is influenced by the wavelength of the light and the geometry of the slits. The distance between adjacent bright or dark bands can be used to derive important parameters such as slit separation and wavelength.

Wavelength and Minima

The wavelength of light is the distance between successive peaks of a wave, and it plays a critical role in determining the positions of minima in an interference pattern. For a double slit setup, the positions of the minima can be calculated using the formula for destructive interference, which involves the wavelength and the slit separation. When comparing two different wavelengths, the relationship between their respective minima can help determine unknown wavelengths based on their positions on the screen.

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