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\(\bullet\) Two slits spaced 0.450 \(\mathrm{mm}\) apart are placed 75.0 \(\mathrm{cm}\) from a screen. What is the distance between the second and third dark lines of the interference pattern on the screen when the slits are illuminated with coherent light with a wavelength of 500 \(\mathrm{nm} ?\)

Short Answer

Expert verified
The distance between the second and third dark lines is approximately 0.833 mm.

Step by step solution

01

Understand the Problem

We need to find the distance between the second and third dark fringes in an interference pattern produced by two slits separated by 0.450 mm, a screen placed 75.0 cm away, and light with a wavelength of 500 nm.
02

Determine the Condition for Dark Fringes

Dark fringes occur at positions where the path difference between light from the two slits is an odd multiple of half the wavelength. The condition for dark fringes is given by \( d \sin \theta = (m + 0.5) \lambda \), where \( d \) is the slit separation, \( \theta \) is the angle relative to the normal of the slits, \( \lambda \) is the wavelength, and \( m \) is the order number.
03

Calculate the Angles for the Second and Third Dark Lines

For the second dark line, \( m = 1 \), so the equation becomes \( 0.450 \times 10^{-3} \sin \theta_2 = (1 + 0.5) \times 500 \times 10^{-9} \). Solve for \( \theta_2 \). For the third dark line, \( m = 2 \), solve \( 0.450 \times 10^{-3} \sin \theta_3 = (2 + 0.5) \times 500 \times 10^{-9} \).
04

Calculate the Positions on the Screen

Using small angle approximation, \( \sin \theta \approx \tan \theta \approx \frac{y}{L} \), where \( y \) is the position on the screen and \( L \) is the distance to the screen. Solve \( y_2 = L \tan \theta_2 \) and \( y_3 = L \tan \theta_3 \), substituting the angles and \( L = 0.75 \) m.
05

Find the Distance Between the Dark Lines

Calculate the distance between the second and third dark lines using \( \Delta y = y_3 - y_2 \).

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Key Concepts

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

Double Slit Experiment
The Double Slit Experiment is a fundamental demonstration of the wave nature of light. When coherent light, such as from a laser, passes through two closely spaced slits, it creates an interference pattern on a screen placed at a distance. This experiment highlights the concept of superposition, where light waves overlap to create regions of constructive (bright) and destructive (dark) interference.

The light from each slit acts as a single wave source. As these waves overlap on the screen, their peaks and troughs interact to produce an interference pattern. This pattern consists of alternating bright and dark bands, known as fringes, which can be explained by the path length difference between the light traveling through each slit. The Double Slit Experiment is critical for understanding wave interference phenomena.
Dark Fringes
Dark fringes occur in the interference pattern of the Double Slit Experiment when waves from the two slits meet out of phase. Instead of reinforcing each other to form a bright band, they cancel out, creating regions of darkness. These are called dark fringes.

The condition for a dark fringe is given by the equation:
  • \[d \sin \theta = (m + 0.5) \lambda\]
where:
  • \(d\) is the distance between the slits,
  • \(\theta\) is the angle relative to the central maximum,
  • \(\lambda\) is the wavelength of the light, and
  • \(m\) is the order of the dark fringe (an integer value).
These parameters help us understand where the dark lines will appear on the screen.
Wavelength of Light
The wavelength of light is a key factor in determining the pattern produced in the Double Slit Experiment. It is the distance between successive peaks or troughs of a wave and is denoted by the Greek letter \(\lambda\).

The wavelength determines the spacing of the interference fringes. In our given exercise, the wavelength of the light used is 500 nm (nanometers), which is a measure of how "long" the wave is. Shorter wavelengths, like blue light, produce more closely spaced fringes, while longer wavelengths, like red light, create wider fringes. Thus, understanding the wavelength is crucial for predicting the position of bright and dark fringes.
Small Angle Approximation
Small Angle Approximation is an important concept in analyzing interference patterns, especially in situations like the Double Slit Experiment. When the angle \(\theta\) is small, we can assume:
  • \(\sin \theta \approx \tan \theta \approx \theta\)
This approach simplifies calculations because \(\theta\), the angle in radians, is approximately equal to \(\sin \theta\) and \(\tan \theta\). For calculating positions of fringes on the screen:
  • \(y \approx L \theta\)
where \(y\) is the fringe position on the screen and \(L\) is the distance to the screen.

In the provided exercise, using this approximation simplifies our task of locating the dark fringes, making it easier to handle without complex trigonometric functions. This is particularly helpful when the screen is far from the slits, ensuring angles remain small.

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Most popular questions from this chapter

\( (f… # \)\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.

\(\bullet\) \(\bullet\) Two speakers that are 15.0 \(\mathrm{m}\) apart produce in- phase sound waves of frequency 250.0 \(\mathrm{Hz}\) in a room where the speed of sound is 340.0 \(\mathrm{m} / \mathrm{s} .\) A woman starts out at the midpoint between the two speakers. The room's walls and ceiling are covered with absorbers to eliminate reflections, and she listens with only one ear for best precision. (a) What does she hear, constructive or destructive interference? Why? (b) She now walks slowly toward one of the speakers. How far from the center must she walk before she first hears the sound reach a minimum intensity? (c) How far from the center must she walk before she first hears the sound maximally enhanced?

\(\bullet\) Electromagnetic waves of wavelength 0.173 nm fall on a crystal surface. As the angle from the plane is gradually increased, starting at \(0^{\circ},\) you find that the first strong interfer- ence maximum occurs when the beam makes an angle of \(22.4^{\circ}\) with the surface of the crystal planes in the Bragg reflection. (a) What is the distance between the crystal planes? (b) At what other angles will interference maxima occur?

\(\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?

\(\bullet\) The lenses of a particular set of binoculars have a coating with index of refraction \(n=1.38,\) and the glass itself has \(n=1.52 .\) If the lenses reflect a wavelength of 525 \(\mathrm{nm}\) the most strongly, what is the minimum thickness of the coating?

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