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Chapter 10: Problem 12

Plane waves of green light \((\lambda=546.1 \mathrm{nm})\) impinge normally on a long narrow slit \((0.15 \mathrm{mm}\) wide ) in an opaque screen. A large lens with a focal length of \(+62.0 \mathrm{cm}\) placed just behind the slit produces a Fraunhofer diffraction pattern on a screen at its focal plane. Determine the width of the central irradiance maximum (zero to zero).

Short Answer

Expert verified
The width of the central maximum is approximately 4.51 mm.

Step by step solution

01

Identify the Given Parameters

Identify and list the given parameters:- Wavelength of the light, \( \lambda = 546.1 \mathrm{nm} = 546.1 \times 10^{-9} \mathrm{m} \).- Width of the slit, \( a = 0.15 \mathrm{mm} = 0.15 \times 10^{-3} \mathrm{m} \).- Focal length of the lens, \( f = 62.0 \mathrm{cm} = 0.62 \mathrm{m} \).
02

Apply the Diffraction Condition

For the central maximum in a single-slit diffraction pattern, the angular positions of the minima are given by the formula:\[ a \sin \theta = m \lambda \]where:- \( a \) is the slit width,- \( \theta \) is the angle of deviation,- \( m \) is the order of the minimum,- \( \lambda \) is the wavelength of the light.For the first minima on either side of the central maximum, \( m = \pm 1 \).
03

Calculate the Angular Position of the First Minima

Set up the equation for the first minimum:\[ 0.15 \times 10^{-3} \sin \theta = 546.1 \times 10^{-9} \]Solve for \( \sin \theta \):\[ \sin \theta = \frac{546.1 \times 10^{-9}}{0.15 \times 10^{-3}} \approx 3.6407 \times 10^{-3} \]
04

Approximating Small Angles

Since the angle \( \theta \) is small, we can use the approximation \( \sin \theta \approx \theta \) in radians. Thus, \( \theta \approx 3.6407 \times 10^{-3} \) radians.
05

Determine the Width of the Central Maximum

The width of the central maximum on the screen can be estimated considering the geometry of the diffraction pattern. The total angular distance between the two first minima is \( 2\theta \), so the linear separation can be calculated as:\[ \text{Width} = 2 \times f \times \theta \]Substituting the known values:\[ \text{Width} = 2 \times 0.62 \times 3.6407 \times 10^{-3} \approx 4.5145 \times 10^{-3} \text{ m} = 4.5145 \text{ mm} \]
06

State the Final Result

Convert the answer to millimeters and state it clearly: The width of the central irradiance maximum is approximately 4.51 mm.

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Key Concepts

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

Single-Slit Diffraction
Single-slit diffraction is a phenomenon that occurs when waves pass through a narrow slit. The waves spread out after passing through the slit, creating an interference pattern. This pattern consists of a series of bright and dark areas known as fringes. These fringes are due to constructive and destructive interference of the waves. In the context of light, these patterns result from light waves spreading out after encountering a slit.
  • The central bright fringe is the widest and most intense.
  • Dark fringes occur where the light waves cancel each other out.
  • Bright fringes reappear at positions where the light waves constructively interfere.
Understanding single-slit diffraction helps explain various optical phenomena and is fundamental in the study of wave optics.
Wavelength of Light
The wavelength of light denotes the distance between successive peaks of a light wave. It is typically measured in nanometers (nm). The wavelength is crucial in determining the color of visible light and its behavior during diffraction.
  • Light with shorter wavelengths, like blue light, diffracts less than longer wavelengths, like red light.
  • The given exercise used a wavelength of 546.1 nm, characteristic of green light.
  • The wavelength directly affects the pattern and spacing of diffracted waves.
Knowing the wavelength allows calculation of diffraction patterns and angles, offering insights into how light moves through spaces.
Diffraction Pattern
A diffraction pattern is the combination of light and dark areas formed by waves spreading and interfering after passing through a slit. In single-slit diffraction, a distinct pattern emerges, centralizing a bright spot surrounded by alternating dark and bright lines.
  • The bright areas are known as maxima and occur where waves enhance each other.
  • Dark areas, or minima, result from wave interference cancelling out light.
  • The structure of a pattern depends on factors such as the slit width and the wavelength of light.
These patterns are significant in multiple fields, like physics and engineering, helping analyze wave behavior in various applications.
Central Maximum
The central maximum refers to the brightest and broadest band in a diffraction pattern. This is where the maximum constructive interference occurs, as light waves align perfectly to enhance brightness.
  • It is flanked by minima where destructive interference takes place, causing darkness.
  • The width of the central maximum increases with a decrease in slit width and decreases with shorter wavelengths.
  • In the original exercise, calculating the width of the central maximum involved determining the angle to the first minima on either side.
Identifying the central maximum is crucial in understanding light behavior in diffraction, providing insights into optical instrument design and function.

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Most popular questions from this chapter

A diffraction grating with slits \(0.60 \times 10^{-3} \mathrm{cm}\) apart is illuminated by light with a wavelength of \(500 \mathrm{nm}\). At what angle will the third-order maximum appear?

Suppose we have 15 parallel long narrow slits in an opaque screen. Furthermore, suppose each slit is separated from the next by a center-to- center distance that is equal to 4 slit widths. Given that a Fraunhofer diffraction pattern appears on a screen, determine the ratio of the irradiance of the second-order principal maximum to that of the zeroth-order maximum.

A transmission grating whose lines are separated by \(3.0 \times\) \(10^{-6} \mathrm{m}\) is illuminated by a narrow beam of red light \(\left(\lambda_{0}=694.3 \mathrm{nm}\right)\) from a ruby laser. Spots of diffracted light, on both sides of the undeflected beam, appear on a screen \(2.0 \mathrm{m}\) away. How far from the central axis is each of the two nearest spots?

If you peered through a 0.75 -mm hole at an eye chart, you would probably notice a decrease in visual acuity. Compute the angular limit of resolution, assuming that it's determined only by diffraction; take \(\lambda_{0}=550 \mathrm{nm} .\) Compare your results with the value of \(1.7 \times 10^{-4} \mathrm{rad}\) which corresponds to a 4.0 -mm pupil.

A narrow single slit (in air) in an opaque screen is illuminated by infrared from a He-Ne laser at \(1152.2 \mathrm{nm}\), and it is found that the center of the tenth dark band in the Fraunhofer pattern lies at an angle of \(6.2^{\circ}\) off the central axis. Determine the width of the slit. At what angle will the tenth minimum appear if the entire arrangement is immersed in water \(\left(n_{w}=1.33\right)\) rather than air \(\left(n_{a}=1.00029\right) ?\)
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