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Red light of wavelength 633 nm from a helium-neon laser passes through a slit 0.350 mm wide. The diffraction pattern is observed on a screen 3.00 m away. Define the width of a bright fringe as the distance between the minima on either side. (a) What is the width of the central bright fringe? (b) What is the width of the first bright fringe on either side of the central one?

Short Answer

Expert verified
Width of central fringe: 10.85 mm. Width of first side fringe: 5.426 mm.

Step by step solution

01

Understanding Diffraction

Light passing through a narrow slit creates a diffraction pattern consisting of bright and dark fringes. The bright fringes occur at positions where the path difference of light waves leads to constructive interference.
02

Calculating the Angles for Minima

The condition for minima in a single-slit diffraction pattern is given by the formula \(a \sin \theta = m\lambda\), where \(a\) is the slit width, \(\lambda\) is the wavelength of light, \(m\) is the order number of the diffraction minima (\(m = \pm 1, \pm 2, ...\)), and \(\theta\) is the angle from the central maximum to the m-th minimum.
03

Finding Minima Positions for the Central Bright Fringe

The central bright fringe is bounded by the first minima on either side ( \(m = \pm 1\)). Calculate \( \theta\) for these by substituting \(\lambda = 633 \, \text{nm} = 633\times10^{-9} \, \text{m}\) and \(a = 0.350\times10^{-3} \, \text{m}\) into \( \sin \theta = \pm \frac{m\lambda}{a}\):\[ \sin \theta_1 = \frac{1 \times 633\times10^{-9}}{0.350\times10^{-3}}\. \]
04

Solving for \( \theta_1 \)

Calculate \( \theta_1 \):\[ \sin \theta_1 = 1.8086\times10^{-3} \]\( \theta_1 \) is small, allowing approximation of \( \theta_1 \approx \sin \theta_1 \).
05

Calculating the Width of the Central Bright Fringe

The width of the central bright fringe is twice the distance from the center to the first minimum. Using \( \theta_1\), calculate the distance \( y_1 \) from the center maximum to the first minimum using \( y_1 = L \tan \theta_1 \approx L \sin \theta_1 \), where \(L = 3.00 \, \text{m}\).\[ y_1 = 3.00 \, \text{m} \cdot 1.8086\times10^{-3} = 5.426 \times10^{-3} \, \text{m} \, \approx 5.426 \, \text{mm} \] Hence, the width of the central fringe is \(2\times5.426\, \text{mm} = 10.85\, \text{mm}\).
06

Finding Minima Positions for First Side Bright Fringe

The first side bright fringe lies between the first minimum (\(m = \pm 1\)) and the second minimum (\(m = \pm 2\)). Calculate \(\sin \theta_2\) for the second minimum:\[ \sin \theta_2 = \frac{2 \times 633\times10^{-9}}{0.350\times10^{-3}}\, \approx 3.6172\times10^{-3} \].
07

Calculating Width of the First Side Bright Fringe

Again, because \( \theta_2\) is small, use approximation \(\theta_2 \approx \sin \theta_2\):Calculate distances \(y_2 = L \sin \theta_2 \approx 3.00 \, \text{m} \cdot 3.6172\times10^{-3}\, \text{m} = 10.85 \, \text{mm}\).Width of first side fringe = \(y_2 - y_1 = 10.85 \, \text{mm} - 5.426 \, \text{mm} = 5.426\, \text{mm}\).

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Key Concepts

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

Diffraction Pattern
When light passes through a narrow slit, it doesn't just continue in a straight line; instead, it spreads out and creates a distinctive pattern on a screen. This is known as a diffraction pattern.
  • A diffraction pattern consists of alternating light and dark regions or fringes.
  • The bright fringes occur where waves interfere constructively.
  • Dark fringes form where there is destructive interference.
The pattern results from the spreading of light waves as they encounter obstacles or openings, leading to a mixture of superimposed waves. These patterns can be used to understand the wave nature of light. In single-slit diffraction, the central light band is the brightest and widest, while subsequent fringes get dimmer and narrower.
Constructive Interference
Interference occurs when two or more wavefronts meet. For bright fringes to appear in a diffraction pattern, constructive interference must happen.
  • Constructive interference is when wavefronts align in such a way that their peaks add up, leading to an increase in light intensity.
  • In terms of the path, lengths from the slit to a particular spot on the screen, the difference in these paths must be multiples of the wavelength.
This alignment is why we see bright bands in a diffraction pattern. For a single-slit setup, bright fringes are spaced where the conditions for constructive interference are met, with the central maximum being the most pronounced due to its alignment with the wavefront.
Central Bright Fringe
The central bright fringe is the most significant feature in a single-slit diffraction pattern. It holds special importance due to its visibility and size.
  • It is located at the center of the pattern, directly opposite the slit.
  • The central bright fringe is the widest and most luminous band in the pattern.
The width of this fringe is defined as the distance between the first minima positions on either side of this bright area. In practice, this means it spans the space where light waves constructively interfere most effectively. It serves as a baseline for comparing other features in the diffraction pattern and helps calculate the wavelength or the slit width.
First Bright Fringe
After the central maximum, we observe additional bright bands known as side fringes. The first bright fringe is the one next to the central fringe.
  • This fringe occurs between the first and second minima.
  • Its width can be calculated by measuring from the center of the slit to these minima positions on the screen.
In this context, its width shows the effect of diffraction beyond the central maximum. The first bright fringe's interaction with light illustrates subsequent points of constructive interference, albeit with decreasing light intensity compared to the central bright fringe. Understanding this fringe is crucial for grasping the underlying principles of wave interference and diffraction in optics.

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Most popular questions from this chapter

Light of wavelength 585 nm falls on a slit 0.0666 mm wide. (a) On a very large and distant screen, how many \(totally\) dark fringes (indicating complete cancellation) will there be, including both sides of the central bright spot? 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 that \(m\) can be?) (b) At what angle will the dark fringe that is most distant from the central bright fringe occur?

A wildlife photographer uses a moderate telephoto lens of focal length 135 mm and maximum aperture \(f/\)4.00 to photograph a bear that is 11.5 m away. Assume the wavelength is 550 nm. (a) What is the width of the smallest feature on the bear that this lens can resolve if it is opened to its maximum aperture? (b) If, to gain depth of field, the photographer stops the lens down to \(f/\)22.0, what would be the width of the smallest resolvable feature on the bear?

Light of wavelength 633 nm from a distant source is incident on a slit 0.750 mm wide, and the resulting diffraction pattern is observed on a screen 3.50 m away. What is the distance between the two dark fringes on either side of the central bright fringe?

If the planes of a crystal are 3.50 \(\AA\) (1 \(\AA\) = 10\(^{-10}\) m = 1 \(\AA\)ngstrom unit) apart, (a) what wavelength of electromagnetic waves is needed so that the first strong interference maximum in the Bragg reflection occurs when the waves strike the planes at an angle of 22.0\(^\circ\), and in what part of the electromagnetic spectrum do these waves lie? (See Fig. 32.4.) (b) At what other angles will strong interference maxima occur?

Parallel rays of green mercury light with a wavelength of 546 nm pass through a slit covering a lens with a focal length of 60.0 cm. In the focal plane of the lens, the distance from the central maximum to the first minimum is 8.65 mm. What is the width of the slit?

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