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Ch. 34 - The Wave Nature of Light: Interference and Polarization
Giancoli Douglas - Physics for Scientists and Engineers 5th edition
Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoGiancoli Douglas 5th editionCh. 34 - The Wave Nature of Light: Interference and PolarizationProblema 2
Capitolo 33, Problema 2

Monochromatic light falling on two slits 0.018 mm apart produces the fifth-order bright fringe at a 12° angle. What is the wavelength of the light used?

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Step 1: Recall the formula for the position of bright fringes in a double-slit interference pattern: \( d \sin \theta = m \lambda \), where \( d \) is the distance between the slits, \( \theta \) is the angle of the bright fringe, \( m \) is the order of the fringe, and \( \lambda \) is the wavelength of the light.
Step 2: Identify the given values from the problem: \( d = 0.018 \; \text{mm} = 0.018 \times 10^{-3} \; \text{m} \), \( \theta = 12^{\circ} \), and \( m = 5 \) (fifth-order bright fringe).
Step 3: Rearrange the formula to solve for the wavelength \( \lambda \): \( \lambda = \frac{d \sin \theta}{m} \).
Step 4: Substitute the known values into the formula: \( \lambda = \frac{(0.018 \times 10^{-3}) \sin(12^{\circ})}{5} \).
Step 5: Use a calculator to compute \( \sin(12^{\circ}) \), multiply by \( d \), and divide by \( m \) to find the wavelength \( \lambda \). Ensure the final result is expressed in meters or converted to nanometers (1 nm = \( 10^{-9} \; \text{m} \)).

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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. When monochromatic light is used, it produces alternating bright and dark fringes on a screen due to constructive and destructive interference, respectively. The position of these fringes depends on the wavelength of the light and the distance between the slits.
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Young's Double Slit Experiment

Interference Pattern

An interference pattern is formed when two or more overlapping waves combine, resulting in regions of increased (constructive interference) and decreased (destructive interference) intensity. In the context of the double-slit experiment, the bright fringes correspond to points where the path difference between the waves from the two slits is an integer multiple of the wavelength, while dark fringes occur at half-integer multiples.
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Wave Interference & Superposition

Wavelength Calculation

The wavelength of light can be calculated using the formula for the position of bright fringes in a double-slit setup: d sin(θ) = mλ, where d is the distance between the slits, θ is the angle of the fringe, m is the order of the fringe, and λ is the wavelength. By rearranging this equation, one can solve for the wavelength when the other variables are known, such as the slit separation and the angle of the observed fringe.
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Unknown Wavelength of Laser through Double Slit
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Domanda del libro di testo

(II) Suppose a thin piece of glass is placed in front of the lower slit in Fig. 34–7 so that the two waves enter the slits 180° out of phase (Fig. 34–44). Describe in detail the interference pattern on the screen.

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Domanda del libro di testo

(II) Light of wavelength λ passes through a pair of slits separated by 0.17 mm, forming a double-slit interference pattern on a screen located a distance 35 cm away. Suppose that the image in Fig. 34–9a is an actual-size reproduction of this interference pattern. Use a ruler to measure a pertinent distance on this image; then utilize this measured value to determine λ (nm) .

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Domanda del libro di testo

(II) Assume that light of a single color, rather than white light, passes through the two-slit setup described in Example 34–4. If the distance from the central fringe to a first-order fringe is measured to be 2.9 mm on the screen, determine the light’s wavelength (in nm) and color (see Fig. 34–11).

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