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Chapter 26: Problem 13

\(\bullet$$\bullet\) Two thin parallel slits that are 0.0116 \(\mathrm{mm}\) apart are illumi- nated by a laser beam of wavelength 585 \(\mathrm{nm}\) . (a) On a very large distant screen, what is the total number of bright fringes (those indicating complete constructive interference), includ- ing the central fringe and those on both sides of it? Solve this problem without calculating all the angles! (Hint: What is the largest that \(\sin \theta\) 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?

Short Answer

Expert verified
(a) 39 bright fringes. (b) The furthest fringe occurs at \(73.7^\circ\).

Step by step solution

01

Understand the Problem

We are given a double-slit experiment with slit separation \(d = 0.0116\, \text{mm} = 1.16 \times 10^{-5}\, \text{m}\) and light wavelength \(\lambda = 585\, \text{nm} = 585 \times 10^{-9}\, \text{m}\). We need to find the number of bright fringes, which occur at angles \(\theta\) satisfying the condition \(d \sin \theta = m \lambda\), where \(m\) is an integer (order number).
02

Determine Maximum Possible Value of \( \sin \theta \)

The maximum possible value for \( \sin \theta \) is 1 (since it is the range of the sine function). Thus,\[ d \sin \theta = m \lambda \leq d \]Substituting given values:\[ m \times 585 \times 10^{-9} \leq 1.16 \times 10^{-5} \]
03

Solve for Maximum \(m\)

Solve the inequality to find the maximum \(m\):\[ m \leq \frac{1.16 \times 10^{-5}}{585 \times 10^{-9}} \]Calculating:\[ m \leq 19.83 \]Thus, the largest integer \(m\) is 19.
04

Count Total Bright Fringes

The possible values for \(m\) range from \(-19\) to 19, including the central fringe at \(m = 0\). Thus, the total number of bright fringes is \(2 \times 19 + 1 = 39\).
05

Calculate the Most Distant Fringe Angle

The most distant fringe corresponds to \(m = \pm 19\). Using the formula:\[ d \sin \theta = m \lambda \]For \(m = 19\):\[ 1.16 \times 10^{-5} \sin \theta = 19 \times 585 \times 10^{-9} \]Solving for \( \theta \),\[ \sin \theta = \frac{19 \times 585 \times 10^{-9}}{1.16 \times 10^{-5}} \approx 0.96 \]\( \theta = \sin^{-1}(0.96) \approx 73.7^\circ \).

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Key Concepts

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

Constructive Interference
In the double-slit experiment, constructive interference occurs when waves from both slits overlap in such a way that their amplitudes add together. This happens at specific angles where the path difference between the light waves is an integer multiple of the light's wavelength.
When this condition is met, the waves combine to create bright spots or fringes on a screen. This process is described by the equation \( d \sin \theta = m \lambda \), where \( d \) is the distance between the slits, \( \theta \) is the angle of the fringe, \( m \) is the fringe order (an integer), and \( \lambda \) is the wavelength of light.
Thus, constructive interference is the principle that results in the bright fringes seen in the double-slit experiment, as the waves amplify each other when their crest and troughs align perfectly.
Fringe Pattern
The fringe pattern in a double-slit experiment refers to the repeating series of bright and dark bands that appear on the screen. These bands are a visual representation of the interference of light waves, created as the light passes through the two slits and spreads out.
- **Bright Fringes:** Occur due to constructive interference when the waves align perfectly.- **Dark Fringes:** Occur due to destructive interference when the waves cancel each other out.
The central bright fringe is typically the most intense as it represents the direct path of the light with no overall path difference, where \( m = 0 \). As we move away from this central fringe, the brightness and spacing of the fringes change according to the angle \( \theta \), which increases along with the order number \( m \).
Wavelength of Light
The wavelength of light, symbolized as \( \lambda \), is the distance between successive peaks of a wave. It is a fundamental property that defines the color of light when visible wavelengths are considered.
In the double-slit experiment, the wavelength is crucial for determining the positions of the interference fringes. The wavelengths can vary across the electromagnetic spectrum, but in the context of the exercise, it is given as 585 nm, which is in the visible region, appearing as a yellow color.
- The equation \( d \sin \theta = m \lambda \) directly uses the wavelength to calculate the angles at which constructive interference occurs.- A longer wavelength would result in more widely spaced fringes, while a shorter wavelength would produce fringes that are closer together.
Interference of Light
Interference of light is a phenomenon that occurs when two or more light waves overlap, resulting in a new wave pattern. This can be either constructive or destructive, depending on how the waves align.
In the double-slit experiment, the primary focus is on interference, as it explains the observed fringe pattern.
  • **Constructive Interference:** Occurs when waves meet in phase, combining to enhance the wave amplitude, shown as bright fringes.
  • **Destructive Interference:** Occurs when waves meet out of phase, canceling each other out, resulting in dark fringes.
The total pattern of light and dark bands is the result of varying angles directly linked to the wavelengths and distances involved. This pattern of interference is not just limited to double-slits, but is a general principle seen in many wave behaviors, illustrating the wave nature of light.

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Most popular questions from this chapter

\(\bullet$$\bullet\) A uniform thin film of material of refractive index 1.40 coats a glass plate of refractive index \(1.55 .\) This film has the proper thickness to cancel normally incident light of wave- length 525 \(\mathrm{nm}\) that strikes the film surface from air, but it is somewhat greater than the minimum thickness to achieve this cancellation. As time goes by, the film wears away at a steady rate of 4.20 nm per year. What is the minimum number of years before the reflected light of this wavelength is now enhanced instead of cancelled?

\(\bullet\) Monochromatic light is at normal incidence on a plane transmission grating. The first-order maximum in the interfer- ence pattern is at an angle of \(8.94^{\circ} .\) What is the angular posi- tion of the fourth-order maximum?

\(\bullet\) A beam of laser light of wavelength 632.8 nm falls on a thin slit 0.00375 \(\mathrm{mm}\) wide. After the light passes through the slit, at what angles relative to the original direction of the beam is it completely cancelled when viewed far from the slit?

\( (f鈥� # \)\bullet$$\bullet\( Two small loudspeakers that are 5.50 \)\mathrm{m}\( apart are emitting sound in phase. From both of them, you hear a singer singing \)\mathrm{C} \\#\( (frequency 277 \)\mathrm{Hz} )\( , while the speed of sound in the room is 340 \)\mathrm{m} / \mathrm{s}\( . Assuming that you are rather far from these speak- ers, if you start out at point \)P\( equidistant from both of them and walk around the room in front of them, at what angles (measured relative to the fine from \)P$ to the midpoint between the speakers) will you hear the sound (a) maximally enhanced, (b) cancelled? Neglect any reflections from the walls.

\(\bullet\) Monochromatic \(x\) rays are incident on a crystal for which the spacing of the atomic planes is 0.440 nm. The first-order maximum in the Bragg reflection occurs when the incident and reflected rays make an angle of \(39.4^{\circ}\) with the crystal planes. What is the wavelength of the rays?
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