Ch 33: Wave Optics

Knight Calc - Physics for Scientists and Engineers 5th Edition

Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook

Knight Calc 5th Edition

Ch 33: Wave Optics

Problem 72

Chapter 33, Problem 72

A double-slit experiment is set up using a helium-neon laser (λ = 633 nm). Then a very thin piece of glass (n = 1.50) is placed over one of the slits. Afterward, the central point on the screen is occupied by what had been the m = 10 dark fringe. How thick is the glass?

Verified step by step guidance

Identify the key concepts: This problem involves the double-slit interference pattern and the effect of introducing a phase shift due to a thin piece of glass. The glass introduces an additional optical path difference, which shifts the interference pattern.

Write the condition for destructive interference: For the central point to now correspond to the m=10 dark fringe, the optical path difference introduced by the glass must equal 10 wavelengths of the laser light. The optical path difference is given by ΔL = (n - 1) * t, where n is the refractive index of the glass and t is its thickness.

Set up the equation for the optical path difference: Since the central point corresponds to the m=10 dark fringe, the optical path difference must satisfy ΔL = m * λ, where m = 10 and λ = 633 nm. Substituting ΔL = (n - 1) * t, we get (n - 1) * t = m * λ.

Rearrange the equation to solve for the thickness of the glass: t = (m * λ) / (n - 1). Substitute the known values: m = 10, λ = 633 nm (convert to meters if needed), and n = 1.50.

Perform the calculation: Plug in the values into the equation t = (m * λ) / (n - 1) to find the thickness of the glass. Ensure the units are consistent (e.g., convert λ to meters if necessary).

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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. When light waves overlap, they can constructively or destructively interfere, resulting in bright and dark fringes on a screen. This experiment is fundamental in understanding quantum mechanics and the dual nature of light.

Phase Change and Optical Path Length

When light passes through different media, its speed changes, affecting its wavelength and phase. A phase change occurs when light reflects off a medium with a higher refractive index. The optical path length, which accounts for both the physical distance and the refractive index of the medium, is crucial for determining the conditions for constructive or destructive interference in the double-slit setup.

Interference and Fringe Order

Interference patterns are characterized by fringe orders, denoted by 'm', which represent the number of wavelengths of path difference between light waves arriving at a point on the screen. Dark fringes occur at specific positions where destructive interference happens, typically at positions given by the formula mλ = d sin(θ), where d is the slit separation and θ is the angle of the fringe. Understanding fringe order is essential for calculating the effects of additional optical elements, like the glass in this scenario.

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Related Practice

Textbook Question

Light of wavelength 600 nm passes though two slits separated by 0.20 mm and is observed on a screen 1.0 m behind the slits. The location of the central maximum is marked on the screen and labeled y = 0. A very thin piece of glass is then placed in one slit. Because light travels slower in glass than in air, the wave passing through the glass is delayed by 5.0×10⁻¹⁶ s in comparison to the wave going through the other slit. What fraction of the period of the light wave is this delay?

FIGURE CP33.74 shows light of wavelength λ incident at angle ϕ on a reflection grating of spacing d. We want to find the angles θ_m at which constructive interference occurs. Light of wavelength 500 nm is incident at ϕ=40° on a reflection grating having 700 reflection lines/mm. Find all angles θ_m at which light is diffracted. Negative values of θ_m are interpreted as an angle left of the vertical.

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Textbook Question

A Michelson interferometer operating at a 600 nm wavelength has a 2.00-cm-long glass cell in one arm. To begin, the air is pumped out of the cell and mirror M₂ is adjusted to produce a bright spot at the center of the interference pattern. Then a valve is opened and air is slowly admitted into the cell. The index of refraction of air at 1.00 atm pressure is 1.00028. How many bright-dark-bright fringe shifts are observed as the cell fills with air?

100

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Textbook Question

FIGURE CP33.73 shows two nearly overlapped intensity peaks of the sort you might produce with a diffraction grating (see Figure 33.9b). As a practical matter, two peaks can just barely be resolved if their spacing Δy equals the width w of each peak, where w is measured at half of the peak’s height. Two peaks closer together than w will merge into a single peak. We can use this idea to understand the resolution of a diffraction grating. In the small-angle approximation, the position of the m = 1 peak of a diffraction grating falls at the same location as the m = 1 fringe of a double slit: y₁= λL/d. Suppose two wavelengths differing by Δλ pass through a grating at the same time. Find an expression for Δy, the separation of their first-order peaks.

1454

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