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\(\bullet\) Light of wavelength 633 nm from a distant source is inci- dent on a slit 0.750 mm wide, and the resulting diffraction pat- tern is observed on a screen 3.50 \(\mathrm{m}\) away. What is the distance between the two dark fringes on either side of the central bright fringe?

Short Answer

Expert verified
The distance between the two dark fringes is approximately 5.90 mm.

Step by step solution

01

Identify Given Information

We need to identify the given information from the problem. The wavelength of light \( \lambda \) is 633 nm, which is equivalent to 633 x 10^(-9) m. The slit width \( a \) is 0.750 mm, or 0.750 x 10^(-3) m. The distance from the slit to the screen \( L \) is 3.50 m.
02

Understand the Concept

In a single-slit diffraction pattern, dark fringes occur at angles \( \theta \) where \( a \sin \theta = m \lambda \), with \( m \) being the order of minimum (here \( m = \pm 1 \) for the first dark fringes on either side of the central maximum).
03

Calculate Angular Position of Dark Fringes

For the first-order dark fringes (\( m = \pm 1 \)), use the formula \( a \sin \theta = m \lambda \). Substitute \( m = 1 \), \( a = 0.750 \times 10^{-3} \) m, and \( \lambda = 633 \times 10^{-9} \) m:\[0.750 \times 10^{-3} \sin \theta = 633 \times 10^{-9}\]Solve for \( \sin \theta \):\[\sin \theta = \frac{633 \times 10^{-9}}{0.750 \times 10^{-3}}\approx 8.44 \times 10^{-4}\]
04

Approximate Small Angle and Find Distance on Screen

Since \( \theta \) is small (diffraction angles are generally small), approximate \( \sin \theta \approx \theta \) (in radians). The linear distance \( y \) from the central bright fringe to the first dark fringe on the screen is given by \( y = L \theta \).\[y = 3.50 \times 8.44 \times 10^{-4} \approx 2.95 \text{ mm}\]
05

Calculate Total Distance Between Dark Fringes

The distance between the first-order dark fringe on one side and the first-order dark fringe on the other side of the central bright fringe is twice \( y \).\[\text{Distance} = 2 \times 2.95 \approx 5.90 \text{ mm}\]

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Key Concepts

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

Wavelength of Light
Light is a form of electromagnetic radiation, and each color of light is distinguished by its wavelength. Wavelength, often denoted by \( \lambda \), is the distance between two consecutive peaks (or troughs) of a wave. For visible light, wavelengths are measured in nanometers (nm), where 1 nm = \(10^{-9}\) meters.
In the given problem, the light wavelength is 633 nm. This wavelength corresponds to a red color in the visible spectrum and is crucial in determining the behavior of the light wave when it encounters obstacles like a slit.
Understanding wavelength helps us predict the diffraction pattern that will result from the interaction of light with obstacles. Shorter wavelengths tend to scatter less compared to longer ones, but all will create diffraction patterns when passing through an aperture like a single slit.
Diffraction Pattern
When light encounters a narrow slit, it bends around the edges, spreading out to produce a pattern called a diffraction pattern. This pattern consists of a series of bright and dark fringes arranged symmetrically around a central bright spot.
The formation of the diffraction pattern can be explained by the principle of wave interference. Waves emerging from different parts of the slit overlap and interfere with each other, resulting in constructive interference (bright fringes) and destructive interference (dark fringes).
These interference patterns are predictable, and the positions of the fringes depend on the slit width, the wavelength of the light, and the distance between the slit and the screen.
Angular Position
The angular position of the dark and bright fringes in a diffraction pattern is determined by the angle \( \theta \). For dark fringes, this angle is where destructive interference occurs, satisfying the equation \( a \sin \theta = m \lambda \).
Here, \( a \) represents the slit width, \( m \) is the order of the minimum which is an integer, and \( \lambda \) is the wavelength. The relationship helps in finding the angles at which the light waves cancel each other out, leading to the dark fringes.
  • \( m \) can take values like \( \pm 1, \pm 2, \pm 3, \ldots \) indicating the first, second, third, etc., order of minima.
  • For small angles, which are typical in diffraction, \( \sin \theta \approx \theta \) in radians, simplifying calculations.
The angular position aids in calculating the actual positions of dark and bright spots on the screen in practical setups.
Linear Distance on Screen
The linear distance of any fringe on the screen from the central bright fringe can be calculated using the formula \( y = L \theta \).
\( L \) is the distance from the slit to the screen, making \( \theta \) small, which is common in many experiments when observing diffraction.
  • This approximation allows us to easily find the distance \( y \), turning circular predictions into linear measurements.
  • It directly connects the angular and spatial requirements of diffraction setups, crucial for interpreting observed patterns.
For example, when calculating the distance between two first-order dark fringes, one on each side of the central bright fringe, you find \( y \) for each side and multiply by two to get the total distance.

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Most popular questions from this chapter

\(\bullet\) Conserving energy. You want to coat the inner surfaces of your windows (which have refractive index of 1.51\()\) with a film in order to enhance the reflection of light back into the room so that you can use bulbs of lower wattage than usual. You find that \(\mathrm{MgF}_{2},\) with \(n=1.38,\) is not too expensive, so you decide to use it. Since incandescent home lightbulbs emit reddish light with a peak wavelength of approximately 650 \(\mathrm{nm}\) , you decide that this wavelength is the one to enhance in the light reflected back into the room. (a) What is the minimum thickness of film that you will need? (b) If this layer seems too thin to be able to put on accurately, what other thicknesses would also work? Give only the three thinnest ones.

\(\bullet$$\bullet\) Two identical audio speakers connected to the same amplifier produce in-phase sound waves with a single fre- quency that can be varied between 300 and 600 Hz. The speed of sound is 340 \(\mathrm{m} / \mathrm{s} .\) You find that where you are standing, you hear minimum-intensity sound. (a) Explain why you hear minimum-intensity sound. (b) If one of the speakers is moved 39.8 \(\mathrm{cm}\) toward you, the sound you hear has maximum inten- sity. What is the frequency of the sound? (c) How much closer to you from the position in part (b) must the speaker be moved to the next position where you hear maximum intensity?

\(\bullet\) Young's experiment is performed with light of wavelength 502 nm from excited helium atoms. Fringes are measured care- fully on a screen 1.20 \(\mathrm{m}\) away from the double slit, and the cen- ter of the 20 th fringe (not counting the central bright fringe) is found to be 10.6 \(\mathrm{mm}\) from the center of the central bright fringe. What is the separation of the two slits?

\(\bullet$$\bullet\) Light of wavelength 585 nm falls on a slit 0.0666 \(\mathrm{mm}\) wide. (a) On a very large distant screen, how many totally dark fringes (indicating complete cancellation) will there be, including both sides of the central bright spot? Solve this prob- lem without calculating all the angles! (Hint: What is the largest that sin \(\theta\) can be? What does this tell you is the largest that \(m\) can be \((\) b) At what angle will the dark fringe that is most distant from the central bright fringe occur?

\(\bullet$$\bullet\) A glass sheet measuring 10.0 \(\mathrm{cm} \times 25.0 \mathrm{cm}\) is covered by a very thin opaque coat- ing. In the middle of this sheet is a thin, straight scratch 0.00125 \(\mathrm{mm}\) thick, as shown in Figure \(26.50 .\) The sheet is totally immersed beneath the surface of a liquid having an index of refraction of \(1.45 .\) Monochromatic light strikes the sheet perpendicular to its surface and passes through the scratch. A screen is placed under water a distance 30.0 \(\mathrm{cm}\) away from the sheet and parallel to it. You observe that the first dark fringes on either side of the central bright fringe on this screen are 22.4 \(\mathrm{cm}\) apart. What is the wavelength of the light in air?

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