A double-slit experiment is to be set up so that the bright fringes appear \(1.27 \mathrm{cm}\) apart on a screen \(2.13 \mathrm{m}\) away from the two slits. The light source was wavelength 500 nm. What should be the separation between the two slits?

Short Answer

Expert verified
The separation between the two slits should be 84 µm, calculated using the double-slit interference formula \(d = \frac{L\lambda}{y}\), with the given values of distance between bright fringes \(y=0.0127\,\mathrm{m}\), distance from the screen \(L=2.13\,\mathrm{m}\), and wavelength of the light source \(\lambda=5\times10^{-7}\,\mathrm{m}\).

Step by step solution

01

Write down the given information

We have the following information: - Distance between bright fringes (y): 1.27 cm - Distance from the screen (L): 2.13 m - Wavelength of light source (λ): 500 nm Before using this information, we need to convert the units as appropriate.
02

Convert units

To solve the problem, we should work with consistent units. Convert cm to m and nm to m. - y = 1.27 cm = 0.0127 m - λ = 500 nm = \(5\times10^{-7}\) m
03

Use the double-slit interference formula

The double-slit interference formula is given by \[y = \frac{L\lambda}{d}\] Where y is the distance between the bright fringes, L is the distance from the screen, λ is the wavelength of the light source, and d is the separation between the slits. Our goal is to find the value of d.
04

Solve for d

Rearrange the formula to solve for d. \[d = \frac{L\lambda}{y}\] Substitute the given values into the formula. \[d = \frac{(2.13\,\mathrm{m})(5\times10^{-7}\,\mathrm{m})}{(0.0127\,\mathrm{m})}\]
05

Calculate the result

Perform the calculation to find the separation between the two slits. \[d =\frac{(2.13)(5\times10^{-7})}{(0.0127)}\] \[d = 8.4\times10^{-5} \,\mathrm{m}\]
06

Convert the result to appropriate units

To express the result in micrometers (µm), convert the result from meters. \[d = 8.4\times10^{-5} \,\mathrm{m} \times \frac{10^6\,\mathrm{\mu m}}{1\,\mathrm{m}} = 84\,\mathrm{\mu m}\] So, the separation between the two slits should be 84 µm.

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Key Concepts

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

Interference Pattern
The double-slit experiment is a classic demonstration of the wave-like properties of light, which shows that light can interfere with itself to produce patterns of light and dark bands known as an interference pattern. This phenomenon occurs when coherent light waves pass through two closely spaced slits and overlap on a distant screen. The bright fringes, or bands, appear where the waves from each slit reinforce one another, also called constructive interference. In contrast, the dark fringes emerge where the waves cancel each other out, known as destructive interference. The pattern's regularity and spacing provide valuable insights into the nature of the light used, including its wavelength.

The positions of these fringes are not arbitrary but relate directly to the light's wavelength, the slit separation, and the distance from the slits to the screen. Understanding the interference pattern is crucial for interpreting the results of the double-slit experiment and for applications in fields such as optics and wave physics.
Wavelength of Light
The wavelength of light is the distance between consecutive peaks or troughs of a wave. It is a fundamental property that determines various characteristics of light, including its color in the visible spectrum. Wavelength is commonly measured in units such as nanometers (nm) for visible light and in other units like meters for different contexts. In our textbook exercise, the wavelength is given as 500 nm, falling within the green portion of the visible spectrum.

When conducting a double-slit experiment, the wavelength governs the spacing of the interference pattern on the screen. The relationship is such that a longer wavelength would result in a wider spacing of the fringes, while a shorter wavelength leads to closer fringes. By analyzing the pattern created, one can deduce the wavelength of the source light if the slit separation and distance to the screen are known. This concept is central in understanding and applying the formula for the double-slit interference pattern.
Slit Separation
The slit separation, denoted as 'd' in the double-slit formula, represents the distance between the two parallel slits in the experiment. This dimension is a critical factor influencing the resulting interference pattern. When the slit separation is comparable to the light's wavelength, the resulting interference effects are readily observable.

In the context of the problem at hand, the slit separation is what we are asked to find, given the wavelength of the light and the distance between the bright fringes on the screen. It is important to realize that the slit separation affects not only the width of the fringes but also their intensity, with narrower slits leading to more pronounced diffraction and thus a more significant impact on the interference pattern. The manipulation of slit separation allows us to control the pattern for various applications or experiments that involve wave properties of light.

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Most popular questions from this chapter

A Michelson interferometer with a He-Ne laser light source \((\lambda=632.8 \mathrm{nm})\) projects its interference pattern on a screen. If the movable mirror is caused to move by \(8.54 \mu \mathrm{m},\) how many fringes will be observed shifting through a reference point on a screen?

Calculate the minimum thickness of an oil slick on water that appears blue when illuminated by white light perpendicular to its surface. Take the blue wavelength to be \(470 \mathrm{nm}\) and the index of refraction of oil to be 1.40.

An inventor notices that a soap bubble is dark at its thinnest and realizes that destructive interference is taking place for all wavelengths. How could she use this knowledge to make a nonreflective coating for lenses that is effective at all wavelengths? That is, what limits would there be on the index of refraction and thickness of the coating? How might this be impractical?

Figure 3.14 shows two glass slides illuminated by monochromatic light incident perpendicularly. The top slide touches the bottom slide at one end and rests on a 0.100-mm-diameter hair at the other end, forming a wedge of air. (a) How far apart are the dark bands, if the slides are \(7.50 \mathrm{cm}\) long and 589 -nm light is used? (b) Is there any difference if the slides are made from crown or flint glass? Explain.

How is the difference in paths taken by two originally in-phase light waves related to whether they interfere constructively or destructively? How can this be affected by reflection? By refraction?

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