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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

Two flat plates of glass with parallel faces are on a table, one plate on the other. Each plate is 11.0 cm long and has a refractive index of 1.55. A very thin sheet of metal foil is inserted under the end of the upper plate to raise it slightly at that end, in a manner similar to that discussed in Example 35.4. When you view the glass plates from above with reflected white light, you observe that, at 1.15 mm from the line where the sheets are in contact, the violet light of wavelength 400.0 nm is enhanced in this reflected light, but no visible light is enhanced closer to the line of contact. (a) How far from the line of contact will green light (of wavelength 550.0 nm) and orange light (of wavelength 600.0 nm) first be enhanced? (b) How far from the line of contact will the violet, green, and orange light again be enhanced in the reflected light? (c) How thick is the metal foil holding the ends of the plates apart?

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?

Red light with wavelength 700 nm is passed through a two-slit apparatus. At the same time, monochromatic visible light with another wavelength passes through the same apparatus. As a result, most of the pattern that appears on the screen is a mixture of two colors; however, the center of the third bright fringe (\(m\) = 32) of the red light appears pure red, with none of the other color. What are the possible wavelengths of the second type of visible light? Do you need to know the slit spacing to answer this question? Why or why not?

Two thin parallel slits that are 0.0116 mm apart are illuminated by a laser beam of wavelength 585 nm. (a) On a very large distant screen, what is the \(total\) number of bright fringes (those indicating complete constructive interference), including the central fringe and those on both sides of it? Solve this problem without calculating all the angles! (\(Hint\): What is the largest that sin u can be? What does this tell you is the largest value of \(m\)?) (b) At what angle, relative to the original direction of the beam, will the fringe that is most distant from the central bright fringe occur?

A plastic film with index of refraction 1.70 is applied to the surface of a car window to increase the reflectivity and thus to keep the car's interior cooler. The window glass has index of refraction 1.52. (a) What minimum thickness is required if light of wavelength 550 nm in air reflected from the two sides of the film is to interfere constructively? (b) Coatings as thin as that calculated in part (a) are difficult to manufacture and install. What is the next greater thickness for which constructive interference will also occur?

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