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A two-slit pattern is viewed on a screen \(1.00 \mathrm{m}\) from the slits. If the two third-order minima are \(22.0 \mathrm{cm}\) apart, what is the width (in \(\mathrm{cm}\) ) of the central bright fringe?

Short Answer

Expert verified
The width of the central bright fringe is 14.67 cm.

Step by step solution

01

Understand the given information

We are told that the two third-order minima are 22.0 cm apart on a screen 1.00 m away from the two slits. We need to find the width of the central bright fringe.
02

Use the formula for minima in a double-slit experiment

The formula for the position of minima in a double-slit pattern is \( y_m = \frac{m \lambda L}{d} \), where \( y_m \) is the position of the minimum, \( m \) is the order number, \( \lambda \) is the wavelength, \( L \) is the distance to the screen, and \( d \) is the slit separation. We have the distance between third-order minima, and we need to use it to find the separation of the central maxima.
03

Express the distance between third-order minima

The distance between two third-order minima is the difference between their positions: \( y_3 - y_{-3} \). Substituting the formula for minima gives \( \frac{3\lambda L}{d} - \left(-\frac{3\lambda L}{d}\right) = \frac{6\lambda L}{d} \). We're given this distance is 22.0 cm.
04

Use the distance between minima to find \( \lambda \) or \( d \)

Since we know \( \frac{6\lambda L}{d} = 0.22 \text{ m} \), we don't have enough direct information yet to solve exactly \( \lambda \) or \( d \) directly, but we can use this equation later to find the central maxima width.
05

Find the width of the central bright fringe

The width of a central bright fringe (central maximum) is given by the first-order maxima, located at \( \pm \frac{\lambda L}{d} \). Thus, the full width (from \( -\) to \( +\)) is \( \frac{2\lambda L}{d} \). Since \( \frac{6\lambda L}{d} = 0.22 \), we have \( \frac{\lambda L}{d} = \frac{1}{3} \times 0.22 \), so \( \frac{2\lambda L}{d} = \frac{2}{3} \times 0.22 = 0.1467 \text{ m} = 14.67 \text{ cm} \).

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Key Concepts

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

Interference Pattern
In the double-slit experiment, light waves passing through two narrow slits will interact with each other, creating an interference pattern on a screen. This pattern is characterized by alternating bright and dark regions, depending on whether the light waves are in-phase (constructive interference) or out-of-phase (destructive interference). When the light waves align perfectly, they amplify each other, forming bright fringes. Conversely, when they cancel each other out, dark fringes are created. The interference pattern is a fundamental demonstration of the wave nature of light, highlighting how light behaves under different conditions. Understanding this pattern is essential in physics as it sets the groundwork for optical technologies and the study of quantum mechanics.
Third-Order Minima
The term "third-order minima" refers to the points in the interference pattern where destructive interference occurs for the third time as you move away from the central maximum. In a double-slit experiment, these minima are the dark fringes that occur due to the condition that the path difference between the two waves equals three wavelengths. Mathematically, it is expressed as: \[ y_m = \frac{m \lambda L}{d} \]where:
  • \(y_m\) is the position of the minimum
  • \(m\) is the order number (in this case, 3 for third-order)
  • \(\lambda\) is the wavelength of the light
  • \(L\) is the distance from the slits to the screen
  • \(d\) is the distance between the slits
The third-order minima are spaced symmetrically about the central bright fringe, and the distance between them tells us about the geometry of the setup, such as the wavelength and the slit separation.
Central Bright Fringe
The central bright fringe, also known as the central maximum, is the brightest spot on the interference pattern directly opposite the slits on the screen. It represents the point of maximum constructive interference where the path difference between the two waves is zero.The position and width of this fringe are crucial in understanding the diffraction pattern. It can be determined using the formula for the first-order maxima:\[ \frac{\lambda L}{d} \] which is then doubled for the full width.This gives a complete span of:\[ \frac{2\lambda L}{d} \]In the given exercise, calculating \( \frac{6\lambda L}{d} = 0.22 \, \text{m}\), allowed us to find the width of the central bright fringe by recognizing that \( \frac{2\lambda L}{d} \) would be \(14.67 \, \text{cm}\). The central bright fringe, being the brightest and most prominent part of the pattern, is often the focal point for measuring these distances and understanding light's wave behavior.

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Most popular questions from this chapter

A diffraction grating with a slit separation \(d\) is illuminated by a beam of monochromatic light of wavelength \(\lambda\). The diffracted beam is observed at an angle \(\phi\) relative to the incident direction. If the plane of the grating bisects the angle between the incident and diffracted beams, show that the \(m\) th maximum will be observed at an angle that satisfies the relation \(m \lambda=2 d \sin (\phi / 2),\) with \(m=0,\pm 1,\pm 2, \ldots\)

A lens that is "optically perfect" is still limited by diffraction effects. Suppose a lens has a diameter of \(120 \mathrm{mm}\) and a focal length of \(640 \mathrm{mm}\). (a) Find the angular width (that is, the angle from the bottom to the top) of the central maximum in the diffraction pattern formed by this lens when illuminated with \(540-\mathrm{nm}\) light. (b) What is the linear width (diameter) of the central maximum at the focal distance of the lens?

In Young's two-slit experiment, the first dark fringe above the central bright fringe occurs at an angle of \(0.31^{\circ} .\) What is the ratio of the slit separation, \(d\), to the wavelength of the light, \(\lambda ?\)

Find the minimum aperture diameter of a camera that can resolve detail on the ground the size of a person \((2.0 \mathrm{m})\) from an SR-71 Blackbird airplane flying at an altitude of \(27 \mathrm{km}\). (Assume light with a wavelength of \(450 \mathrm{nm}\).)

Is resolution greater with blue light or red light, all other factors being equal? Explain.

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