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Interference Fringes A double-slit arrangement produces interference fringes for sodium light \((\lambda=589 \mathrm{~nm})\) that have an angular separation of \(3.50 \times 10^{-3}\) rad. For what wavelength would the angular separation be \(10.0 \%\) greater?

Short Answer

Expert verified
The wavelength for a 10.0% greater angular separation is 647.9 nm.

Step by step solution

01

- Understand the relationship

The angular separation \theta of the fringes in a double-slit experiment is given by the formula: \[ \theta = \frac{\beta}{D} = \frac{\frac{\rho \times \theta}{d}}{D} \] where \( \theta \) is the angular separation, \( \beta \) is the fringe width, \( D \) is the distance between the slits, \( \rho \) is the distance from the slits to the screen, and \( d \) is the distance between the slits. Simplified, \[ \theta = \frac{\rho \times \theta}{d} \].
02

- Calculate the initial angular separation

The initial angular separation is given as \(3.50 \times 10^{-3}\) rad and corresponds to the wavelength \( \rho = 589 \text{ nm} \) which is suitable.
03

- Determine the new angular separation

The new angular separation would be increased by \(10\text{%}\): \[ \theta_{\text{new}} = 1.10 \times \theta = 1.10 \times 3.50 \times 10^{-3} \text{ rad} \] Solve this step.
04

- Simplify and solve for new wavelength

Since angular separation is proportional to wavelength, we use the relationship: \[ \theta_{\text{new}} = \frac{\rho_{\text{new}}}{d} \] So the updated formula since \( d\) remains constant: \[ \rho_{\text{new}} = \rho + (0.1 \times \rho) = 1.10 \times \rho = 1.10 \times 589 \text{ nm} \]
05

- Calculate the final result

Compute: \( \rho_{\text{new}} = 1.10 \times 589 \text{ nm} = 647.9 \text{ nm}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

angular separation
The term 'angular separation' refers to the angle between consecutive interference fringes produced in a double-slit experiment. When light passes through two closely spaced slits, it creates a pattern on a screen due to the wave nature of light. These patterns are called interference fringes. The formula to calculate the angular separation of these fringes is:
\[ \theta = \frac{\beta}{D} = \frac{\frac{\rho \times \theta}{d}}{D} \]
Here:
  • θ is the angular separation.
  • β is the fringe width.
  • ÒÏ is the distance from the slits to the screen.
  • d is the distance between the slits.

Understanding this relationship helps us change variables such as wavelength to alter the angular separation. For example, if we need to increase the angular separation by 10%, we adjust the wavelength accordingly since angular separation is directly proportional to wavelength.
wavelength calculation
Wavelength is a key factor in determining the interference pattern in a double-slit experiment. In our given problem, we know the original wavelength of the sodium light is 589 nm. Angular separation is proportional to wavelength, given the equation: \[ \theta_{\text{new}} = 1.10 \times \theta \]
This relationship enables us to find the new wavelength when the angular separation is increased by 10%. Thus, the new wavelength:
\[ \rho_{\text{new}} = 1.10 \times \rho \]
By inserting the values, we get:
\[ \rho_{\text{new}} = 1.10 \times 589 \text{ nm} = 647.9 \text{ nm} \]
This calculation shows that, by increasing the angular separation by 10%, the new wavelength required is 647.9 nm.
interference patterns
Interference patterns are a fundamental observation in wave optics, specifically in experiments like the double-slit experiment. When coherent light waves pass through closely spaced slits, they superpose and create bright and dark fringes on a screen. These bright and dark regions result from constructive and destructive interferences, respectively.

  • Constructive Interference: When the paths are equal to a whole number of wavelengths, bright fringes form.
  • Destructive Interference: When paths differ by half a wavelength, dark fringes form.

These patterns provide insight into the wave nature of light and help in measuring wavelengths and slit separations. Angular separation plays a key role in determining the distance between these fringes, and understanding how to manipulate this can reveal various properties of the waves.
wave optics
Wave optics, or physical optics, studies the behavior of light as a wave. This field covers phenomena such as interference, diffraction, and polarization, which cannot be explained by geometric optics alone.
The double-slit experiment is one of the classic examples of wave optics, demonstrating the wave nature of light. When studying wave optics:
  • Interference: This is when two or more wavefronts overlap, resulting in a new wave pattern.
  • Diffraction: This occurs when a wave encounters an obstacle or slit and bends around it, creating a new wave pattern.
  • Polarization: This is the orientation of vibrations in the wave, especially in transverse waves like light.

In double-slit experiments and other wave optics phenomena, the wavelength of light significantly impacts the observed patterns, such as the angular separation of interference fringes. By altering wavelength, we can directly influence these wave patterns, making wave optics a valuable tool in various scientific and industrial applications.

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Most popular questions from this chapter

Soap Film Two The reflection of perpendicularly incident white light by a soap film in air has an interference maximum at \(600 \mathrm{~nm}\) and a minimum at \(450 \mathrm{~nm}\), with no minimum in between. If \(n=\) \(1.33\) for the film, what is the film thickness, assumed uniform?

Mica Flake A thin flake of mica \((n=1.58)\) is used to cover one slit of a double-slit interference arrangement. The central point on the viewing screen is now occupied by what had been the seventh bright side fringe \((m=7)\) before the mica was used. If \(\lambda=550 \mathrm{~nm}\), what is the thickness of the mica? (Hint: Consider the wavelength of the light within the mica.)

Newtons's Rings One In a Newton's rings experiment (see Problem 49), the radius of curvature \(R\) of the lens is \(5.0 \mathrm{~m}\) and the lens diameter is \(20 \mathrm{~mm}\). (a) How many bright rings are produced? Assume that \(\lambda=589 \mathrm{~nm}\). (b) How many bright rings would be produced if the arrangement were immersed in water \((n=1.33)\) ?

Soap Film Light of wavelength \(624 \mathrm{~nm}\) is incident perpendicularly on a soap film (with \(n=1.33\) ) suspended in air. What are the least two thicknesses of the film for which the reflections from the film undergo fully constructive interference?

Directly Downward In Fig. \(36-25\), white light is sent directly downward through the top plate of a pair of glass plates. The plates touch at the left end and are separated by a wire of diameter \(0.048\) \(\mathrm{mm}\) at the right end; the air between the plates acts as a thin film. An observer looking down through the top plate sees bright and dark fringes due to that film. (a) Is a dark fringe or a bright fringe seen at the left end? (b) To the right of that end, fully destructive interference occurs at different locations for different wavelengths of the light. Does it occur first for the red end or the blue end of the visible spectrum?

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