Problem 15 - University Physics with Modern Physics

Question

Coherent light with wavelength 600 nm passes through two very narrow slits and the interference pattern is observed on a screen 3.00 m from the slits. The first-order bright fringe is at 4.84 mm from the center of the central bright fringe. For what wavelength of light will the first-order dark fringe be observed at this same point on the screen?

Step-by-Step Solution

Answer

The wavelength for the first-order dark fringe at 4.84 mm is 400 nm.

Step 1: Understanding the Problem

We have a double-slit interference pattern created by light of a known wavelength of 600 nm, and we observe the first-order bright fringe at a distance of 4.84 mm from the center. We need to find a wavelength that results in a first-order dark fringe at the same distance from the center.

Step 2: Formula for Bright and Dark Fringes

For the first-order bright fringe, the position is given by \( m \lambda = d \frac{x}{L} \), where \( m \) is the order (1 for first-order), \( \lambda \) is the wavelength, \( d \) is the distance between the slits, \( x \) is the fringe position, and \( L \) is the distance to the screen. For the first-order dark fringe, the formula is \( (m + \frac{1}{2}) \lambda = d \frac{x}{L} \).

Step 3: Finding the Distance Between the Slits, d

Using the given wavelength 600 nm and the position of the bright fringe, we solve for \( d \) using \( 600 \times 10^{-9} \times 1 = d \frac{4.84 \times 10^{-3}}{3} \). Solving for \( d \), we find \( d = \frac{600 \times 10^{-9} \times 3}{4.84 \times 10^{-3}} \approx 3.72 \times 10^{-4} \) m.

Step 4: Applying Formula for the Dark Fringe

Use the formula for the dark fringe \( (1 + \frac{1}{2}) \lambda = d \frac{4.84 \times 10^{-3}}{3} \). Substitute \( d = 3.72 \times 10^{-4} \) m to find the new wavelength \( \lambda \).

Step 5: Solving for New Wavelength

Rearrange to solve for \( \lambda \), \( \lambda = \frac{2}{3} \times \frac{3.72 \times 10^{-4} \times 3}{4.84 \times 10^{-3}} \). This equals \( \approx 400 \times 10^{-9} \) m or 400 nm.

Key Concepts

Double-slit experiment

The double-slit experiment is a classic demonstration of the wave nature of light. When coherent light, such as a laser, passes through two closely spaced slits, it creates an interference pattern on a screen placed behind these slits. This pattern is characterized by a series of alternating light and dark bands, known as fringes.

The light waves emerging from the slits overlap and interact with each other. Where the waves meet in phase, meaning their peaks align, they constructively interfere, creating bright fringes. Conversely, where the waves are out of phase, their peaks and troughs cancel out, resulting in dark fringes. This experiment highlights how light behaves both as a particle and a wave, a central tenet of quantum mechanics. Understanding this experiment contributes to insights into diffraction and the wave-particle duality of light.

• Basic principle: Interference of light waves
• Requirements: Coherent light source and two narrow slits
• Outcome: Pattern of bright and dark fringes

Wavelength calculation

Calculating the wavelength of light involves understanding the mathematical relationships governing interference patterns. In the double-slit experiment, the position of bright and dark fringes is related to the wavelength of the light passing through the slits. The formula used for calculating the bright fringe positions is \( m \lambda = d \frac{x}{L} \), where \( m \) is the order of the fringe, \( \lambda \) is the light wavelength, \( d \) is the slit separation, \( x \) is the fringe's position on the screen, and \( L \) is the distance from the slits to the screen.

To find a wavelength that creates a dark fringe at a specific position, the formula is slightly altered to incorporate a phase difference for destructive interference: \( \left(m + \frac{1}{2}\right) \lambda = d \frac{x}{L} \). These formulas allow us to determine unknown quantities such as the slit separation or a new wavelength causing dark fringes at the same location as previously observed bright fringes.

Dark and bright fringes

The pattern observed on the screen in the double-slit experiment consists of alternating dark and bright fringes. These are visible evidences of the interference of light waves emanating from the slits. The bright fringes occur due to constructive interference where the path difference of the overlapping light waves is a multiple of the wavelength. For instance, at the first-order bright fringe, this path difference equals one wavelength.

Dark fringes, on the other hand, arise from destructive interference. Here, the path difference is an odd multiple of half wavelengths. This condition ensures that the peak of one wave coincides with the trough of another, leading to cancellation. The detailed patterns are determined by the wavelength of the light, the slit distance, and the distance to the observation screen. Understanding these patterns helps in exploring fundamental physical phenomena like wave interference and diffraction.