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Chapter 36: Problem 15

A slit 0.240 \(\mathrm{mm}\) wide is illuminated by parallel light rays of wavelength 540 \(\mathrm{nm}\) . The diffraction pattern is observed on a screen that is 3.00 \(\mathrm{m}\) from the slit. The intensity at the center of the central maximum \(\left(\theta=0^{\circ}\right)\) is \(6.00 \times 10^{-6} \mathrm{W} / \mathrm{m}^{2} .\) (a) What is the distance on the screen from the center of the central maximum to the first minimum? (b) What is the intensity at a point on the screen midway between the center of the central maximum and the first minimum?

Short Answer

Expert verified
(a) 6.75 mm; (b) 2.43 脳 10^-6 W/m^2.

Step by step solution

01

Identify Parameters for the Problem

First, gather the key parameters for the problem: the slit width is given as \( a = 0.240 \, \text{mm} = 0.240 \times 10^{-3} \; \text{m} \), the wavelength is \( \lambda = 540 \; \text{nm} = 540 \times 10^{-9} \; \text{m} \), and the distance to the screen is \( L = 3.00 \; \text{m} \). The intensity at the center of the central maximum is \( I_0 = 6.00 \times 10^{-6} \, \text{W/m}^2 \).
02

Determine the Condition for the First Minimum

The angular position \( \theta \) for the first minimum in the diffraction pattern (single-slit diffraction) is given by the formula \( a \sin\theta = m\lambda \), where \( m \) is the order of the minimum. For the first minimum, \( m = 1 \). Thus, \( 0.240 \times 10^{-3} \sin\theta = 1 \times 540 \times 10^{-9} \). Solve for \( \sin\theta \).
03

Calculate the Angle for the First Minimum

Solve the equation for \( \sin\theta \):\[\sin\theta = \frac{540 \times 10^{-9}}{0.240 \times 10^{-3}} = 2.25 \times 10^{-3}.\]Since \( \theta \) is very small, we can use \( \sin\theta \approx \theta \) in radians. Hence, \( \theta \approx 2.25 \times 10^{-3} \; \text{radians} \).
04

Calculate the Distance to the First Minimum

Use the small angle approximation to find the distance \( y \) on the screen: \( y = L \tan\theta \approx L \sin\theta \) (since \( \tan\theta \approx \sin\theta \) for small angles). Therefore,\[y = 3.00 \times 2.25 \times 10^{-3} = 6.75 \times 10^{-3} \; \text{m} \text{ or } 6.75 \; \text{mm}.\]
05

Determine the Intensity Midway to the First Minimum

The intensity at an angle \( \theta \) for a single slit is given by the formula:\[I(\theta) = I_0 \left(\frac{\sin\beta}{\beta}\right)^2,\]where \( \beta = \frac{\pi a \sin\theta}{\lambda} \). Halfway to the first minimum indicates \( \theta_{mid} = \frac{2.25 \times 10^{-3}}{2} = 1.125 \times 10^{-3} \; \text{radians} \).
06

Calculate the Intensity at \( \theta_{mid} \)

First calculate \( \beta_{mid} = \frac{\pi \times 0.240 \times 10^{-3} \times 1.125 \times 10^{-3}}{540 \times 10^{-9}} \) which simplifies to \( \beta_{mid} = 1.57 \). Then,\[I(\theta_{mid}) = I_0 \left(\frac{\sin(1.57)}{1.57}\right)^2 \approx 6.00 \times 10^{-6} \left(\frac{1}{1.57}\right)^2 \approx 2.43 \times 10^{-6} \; \text{W/m}^2.\]This represents the intensity at the midpoint between the center and first minimum.

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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 occurs when light passes through a narrow slit and spreads out, forming a pattern of light and dark regions. This phenomenon is rooted in Huygens' principle, which states that every point on a wavefront acts as a source of secondary wavelets. As these waves spread out from the slit, they overlap and interfere with each other. This is known as interference, which leads to the formation of diffraction patterns on a screen.

In single slit diffraction, each part of the slit acts like a point source, and the light waves interfere both constructively and destructively. Constructive interference leads to bright fringes, known as maxima, while destructive interference creates dark fringes, known as minima.
  • The position of minima is defined by the angle, \( \theta \), where the destructive interference results in darkness.
  • Using the formula \( a \sin\theta = m\lambda \), where \( a \) is the slit width and \( m \) is the order of the minimum, you can calculate the position of the minima.
This simple setup reveals complex and beautiful principles of wave behavior, allowing us to determine characteristics like the wavelength of light.
Diffraction Pattern
A diffraction pattern is the series of light and dark fringes observed on a screen as a result of wave interference, specifically in scenarios like single slit diffraction. When light waves pass through a slit, they bend and spread out, leading to interference, which forms these patterns.

For a single slit, the central bright fringe, or maximum, is the most intense and widest part of the diffraction pattern. This is flanked by alternating dark and bright fringes. Each dark fringe corresponds to a point of destructive interference, where waves cancel each other out, producing a minimum in intensity.
  • The spacing and intensity of these fringes help us understand the interference and diffraction mechanisms.
  • The central fringe, being the brightest and broadest, contains most of the light's energy and is located at \( \theta = 0^\circ \).
The formula \( I(\theta) = I_0 \left( \frac{\sin\beta}{\beta} \right)^2 \) is used to describe the intensity at any angle \( \theta \), showcasing how light spreads through the slit and onto a screen, creating this pattern.
Wavelength
The wavelength of light, often denoted by the symbol \( \lambda \), is a crucial factor in diffraction phenomena. Wavelength is the distance between two consecutive peaks of a wave and determines its color in the spectrum of visible light.

When light passes through a narrow slit, its wavelength interacts with the slit width to determine the diffraction pattern formed on a screen. The wavelength, in conjunction with the slit width, defines the interference conditions which create patterns of light and shadow.
  • The greater the wavelength compared to slit width, the more pronounced the diffraction effects and fewer fringes produced.
  • In single-slit diffraction, the wavelength is a key part of the equation \( a \sin\theta = m\lambda \), determining where minima and maxima will occur.
Understanding wavelength allows us to predict how light interacts with obstacles, enhancing our comprehension of wave behavior in various contexts, including technology utilizing lasers or holography.

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Most popular questions from this chapter

Monochromatic light of wavelength 580 nm passes through a single slit and the diffraction pattern is observed on a screen. Both the source and screen are far enough from the slit for Fraunhofer diffraction to apply.(a) If the first diffraction minima are at \(\pm 90.0^{\circ},\) so the central maximum completely fills the screen, what is the width of the slit? (b) For the width of the slit as calculated in part (a), what is the ratio of the intensity at \(\theta=45.0^{\circ}\) to the intensity at \(\theta=0 ?\)

Laser light of wavelength 632.8 nm falls normally on a slit that is 0.0250 \(\mathrm{mm}\) wide. The transmitted light is viewed on a distant screen where the intensity at the center of the central bright fringe is 8.50 \(\mathrm{W} / \mathrm{m}^{2} .\) (a) Find the maximum number of totally dark fringes on the screen, assuming the screen is large enough to show them all. (b) At what angle does the dark fringe that is most distant from the center occur? (c) What is the maximum intensity of the bright fringe that occurs immediately before the dark fringe in part (b)? Approximate the angle at which this fringe occurs by assuming it is midway between the angles to the dark fringes on either side of it.

Monochromatic light of wavelength \(\lambda=620 \mathrm{nm}\) from a distant source passes through a slit 0.450 \(\mathrm{mm}\) wide. The diffraction pattern is observed on a screen 3.00 \(\mathrm{m}\) from the slit. In terms of the intensity \(I_{0}\) at the peak of the central maximum, what is the intensity of the light at the screen the following distances from the center of the central maximum: (a) \(1.00 \mathrm{mm} ;\) (b) 3.00 \(\mathrm{mm}\) ; (c) 5.00 \(\mathrm{mm}\) ?

An interference pattern is produced by light of wave- length 580 \(\mathrm{nm}\) from a distant source incident on two identical parallel slits separated by a distance (between centers) of 0.530 \(\mathrm{mm}\) . (a) If the slits are very narrow, what would be the angular positions of the first-order and second- order, two-slit, interference maxima? (b) Let the slits have width 0.320 \(\mathrm{mm}\) . In terms of the intensity \(I_{0}\) at the center of the central maximum, what is the intensity at each of the angular positions in part (a)?

Searching for Starspots. The Hale Telescope on Palomar Mountain in California has a mirror 200 in. \((5.08 \mathrm{m})\) in diameter and it focuses visible light. Given that a large sunspot is about \(10,000\) mi in diameter, what is the most distant star on which this telescope could resolve a sunspot to see whether other stars have them? (Assume optimal viewing conditions, so that the resolution is diffraction limited.) Are there any stars this close to us, besides our sun?
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