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Coherent light with wavelength 450 nm falls on a pair of slits. On a screen 1.80 m away, the distance between dark fringes is 3.90 mm. What is the slit separation?

Short Answer

Expert verified
The slit separation is approximately 2.08 x 10^{-4} m.

Step by step solution

01

Understand the Concept

The problem involves two-slit interference. The formula for the position of dark fringes in a double-slit experiment is given by: \[ d \sin \theta = (m + \frac{1}{2}) \lambda \] where \( d \) is the slit separation, \( \theta \) is the angle of the fringe, \( m \) is the order number, and \( \lambda \) is the wavelength of light. The distance between fringes on the screen can also be related to the geometry of the setup.
02

Identify Known Values

We are given:- Wavelength, \( \lambda = 450 \) nm = \( 450 \times 10^{-9} \) m- Distance between the screen and the slits, \( L = 1.80 \) m- Distance between dark fringes, \( \Delta y = 3.90 \) mm = \( 3.90 \times 10^{-3} \) mThese values will be used in calculations to determine the slit separation.
03

Relate Fringe Spacing to Given Values

For small angles, \( \sin \theta \approx \tan \theta \times \theta \approx \frac{y}{L} \). The distance between two consecutive dark fringes (fringe spacing) is given by:\[ \Delta y = \frac{\lambda L}{d} \]This formula relates the fringe spacing, wavelength, slit separation, and distance between the screen and slits.
04

Solve for Slit Separation

Re-arranging the equation \( \Delta y = \frac{\lambda L}{d} \) for \( d \), we have:\[ d = \frac{\lambda L}{\Delta y} \]Substitute the known values:\[ d = \frac{450 \times 10^{-9} \times 1.80}{3.90 \times 10^{-3}} \]
05

Calculate the Result

Carrying out the calculation:\[ d = \frac{450 \times 10^{-9} \times 1.80}{3.90 \times 10^{-3}} = \frac{810 \times 10^{-9}}{3.90 \times 10^{-3}} \approx 2.08 \times 10^{-4} \text{ m} \]Thus, the slit separation \( d \) is approximately \( 2.08 \times 10^{-4} \) meters.

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Key Concepts

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

Wavelength
In the context of the double-slit interference experiment, the wavelength is a key component that describes the distance between consecutive peaks (or troughs) of a wave. Wavelength is usually denoted by the Greek letter \( \lambda \) and is often measured in nanometers (nm) for visible light.

The wavelength in this exercise is 450 nm, equivalent to \( 450 \times 10^{-9} \) meters. This value indicates the color of the light, which in this case, is in the blue spectrum. Different wavelengths will affect how light interferes with itself, changing the pattern of light and dark bands observed on the screen. Shorter wavelengths will typically lead to closer spacing of interference fringes, while longer wavelengths result in wider spacing.

Understanding the wavelength allows us to predict the behavior of light in various interference setups, such as through different slit separations or varying distances to the screen.
Fringe Spacing
Fringe spacing refers to the distance between two consecutive bright or dark fringes on the screen in a double-slit experiment. This distance, often denoted by \( \Delta y \), is determined by the interaction of coherent light waves as they superimpose on one another after passing through the slits.

In this exercise, the fringe spacing is 3.90 mm, or \( 3.90 \times 10^{-3} \) m. It represents how the interference pattern spreads out on the screen. Larger fringe spacing occurs when the interference pattern is spread more widely, which can be the result of varying factors such as the slit separation, the wavelength of the light, or the distance to the screen.
  • Formulas: The relationship between fringe spacing and other factors can be expressed as \( \Delta y = \frac{\lambda L}{d} \). This equation considers the wavelength \( \lambda \), slit separation \( d \), and the screen distance \( L \).
  • Factors Influencing Fringe Spacing: Changes in wavelength or slit separation have a direct impact on \( \Delta y \), altering the interference pattern observed.
Slit Separation
Slit separation, denoted as \( d \), is an essential factor in determining the pattern created in a double-slit experiment. The distance between the two slits affects how the light waves emerge and interfere with each other.

In this exercise, solving for the slit separation involves using the formula \( d = \frac{\lambda L}{\Delta y} \). By substituting the known values of wavelength (450 nm), screen distance (1.80 m), and fringe spacing (3.90 mm), the slit separation is found to be approximately \( 2.08 \times 10^{-4} \) meters.
  • As the slit separation \( d \) decreases, the interference pattern generally becomes more spread out, leading to wider fringe spacing.
  • A larger slit separation causes the fringes to appear closer together on the screen.
  • Understanding and calculating slit separation is crucial for predicting and manipulating the interference pattern of waves.
Screen Distance
The distance from the slits to the screen, denoted as \( L \), is a critical factor in determining the size and distribution of the interference pattern in a double-slit experiment. This parameter influences how the light waves overlap after passing through the slits.

For this problem, the screen distance is given as 1.80 meters. An increase in this distance tends to spread the fringes farther apart, leading to larger fringe spacing. Conversely, decreasing the screen distance would bring the fringes closer together, decreasing fringe spacing.
  • Geometric Relationship: The angle \( \theta \) at which lights interfere is small, so \( \sin \theta \approx \tan \theta \approx \frac{y}{L} \).
  • Impact on Pattern: Changes in the screen distance directly affect the interference spacing, thus influencing which order of fringes might be visible.
Understanding screen distance and its relationship with other factors such as wavelength, fringe spacing, and slit separation helps in designing experiments to observe specific interference patterns.

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Most popular questions from this chapter

Suppose you illuminate two thin slits by monochromatic coherent light in air and find that they produce their first interference \(minima\) at \(\pm35.20^\circ\) on either side of the central bright spot. You then immerse these slits in a transparent liquid and illuminate them with the same light. Now you find that the first minima occur at \(\pm19.46^\circ\) instead. What is the index of refraction of this liquid?

A radio transmitting station operating at a frequency of 120 MHz has two identical antennas that radiate in phase. Antenna B is 9.00 m to the right of antenna A. Consider point P between the antennas and along the line connecting them, a horizontal distance x to the right of antenna A. For what values of x will constructive interference occur at point P?

Laser light of wavelength 510 nm is traveling in air and shines at normal incidence onto the flat end of a transparent plastic rod that has \(n\) = 1.30. The end of the rod has a thin coating of a transparent material that has refractive index 1.65. What is the minimum (nonzero) thickness of the coating (a) for which there is maximum transmission of the light into the rod; (b) for which transmission into the rod is minimized?

In your research lab, a very thin, flat piece of glass with refractive index 1.40 and uniform thickness covers the opening of a chamber that holds a gas sample. The refractive indexes of the gases on either side of the glass are very close to unity. To determine the thickness of the glass, you shine coherent light of wavelength \(\lambda_0\) in vacuum at normal incidence onto the surface of the glass. When \(\lambda_0\) = 496 nm, constructive interference occurs for light that is reflected at the two surfaces of the glass. You find that the next shorter wavelength in vacuum for which there is constructive interference is 386 nm. (a) Use these measurements to calculate the thickness of the glass. (b) What is the longest wavelength in vacuum for which there is constructive interference for the reflected light?

Two small stereo speakers \(A\) and \(B\) that are 1.40 m apart are sending out sound of wavelength 34 cm in all directions and all in phase. A person at point \(P\) starts out equidistant from both speakers and walks so that he is always 1.50 m from speaker \(B\) (Fig. E35.1). For what values of x will the sound this person hears be (a) maximally reinforced, (b) cancelled? Limit your solution to the cases where x \(\leq\) 1.50 m.

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