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Chapter 36: Problem 21

An interference pattern is produced by light of wavelength 580 nm from a distant source incident on two identical parallel slits separated by a distance (between centers) of 0.530 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 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)?

Short Answer

Expert verified
First-order maxima angle \( \theta_1 \approx 0.0627^\circ \) and second-order maxima angle \( \theta_2 \approx 0.1254^\circ \). Intensities depend on calculated beta values for each angle.

Step by step solution

01

Understand the Problem

We are given a two-slit interference problem with a light source of wavelength 580 nm. The slits are separated by 0.530 mm, and we need to find the angular positions of the interference maxima as well as the intensity at those angles.
02

First-order Interference Maxima

The condition for constructive interference in a two-slit experiment is given by \( d \sin \theta = m \lambda \), where \( d \) is the distance between the slits, \( \theta \) is the angle of interference, \( m \) is the order of the maximum, and \( \lambda \) is the wavelength of light. For the first order \( (m=1) \), we have:\[ \sin \theta_1 = \frac{m \lambda}{d} = \frac{1 \times 580 \times 10^{-9}\, \text{m}}{0.530 \times 10^{-3}\, \text{m}} \]\[ \sin \theta_1 \approx 0.001094 \]Therefore, \( \theta_1 \approx \arcsin(0.001094) \).
03

Second-order Interference Maxima

Apply the same formula for \( m=2 \):\[ \sin \theta_2 = \frac{2 \times 580 \times 10^{-9}\, \text{m}}{0.530 \times 10^{-3}\, \text{m}} \]\[ \sin \theta_2 \approx 0.002188 \]Therefore, \( \theta_2 \approx \arcsin(0.002188) \).
04

Calculate Angular Positions

Using a calculator, compute \( \theta_1 \approx \arcsin(0.001094) \approx 0.0627^\circ \) and \( \theta_2 \approx \arcsin(0.002188) \approx 0.1254^\circ \).
05

Consider Slit Width Influence on Intensity

Next, we calculate the intensity at each angular position considering the slit width of 0.320 mm. The intensity formula modified by the single-slit diffraction pattern is:\[ I(\theta) = I_0 \left(\frac{\sin(\beta)}{\beta}\right)^2 \]where \( \beta = \frac{\pi a \sin \theta}{\lambda} \) and \( a = 0.320 \times 10^{-3}\, \text{m} \).
06

Intensity at First-order Maximum

For the first-order maximum, \( \theta = \theta_1 \):\[ \beta_1 = \frac{\pi \times 0.320 \times 10^{-3} \times \sin(0.0627^\circ)}{580 \times 10^{-9}} \]Calculate \( \sin(\beta_1)/\beta_1 \) and then find \( I(\theta_1) = I_0 \left(\frac{\sin(\beta_1)}{\beta_1}\right)^2 \).
07

Intensity at Second-order Maximum

For the second-order maximum, \( \theta = \theta_2 \):\[ \beta_2 = \frac{\pi \times 0.320 \times 10^{-3} \times \sin(0.1254^\circ)}{580 \times 10^{-9}} \]Calculate \( \sin(\beta_2)/\beta_2 \) and then find \( I(\theta_2) = I_0 \left(\frac{\sin(\beta_2)}{\beta_2}\right)^2 \).
08

Evaluate Final Intensities

With the calculations of \( \beta_1 \) and \( \beta_2 \), insert the values into the respective intensity formulas to yield the intensity proportions relative to \( I_0 \).

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Key Concepts

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

Two-Slit Experiment
The two-slit experiment, famously conducted by Thomas Young, illustrates the wave nature of light through the phenomenon of interference. When coherent light, meaning light of constant phase difference like that from a laser, passes through two close, narrow slits, it creates a pattern of alternating bright and dark fringes on a screen. These patterns result from constructive and destructive interference of light waves.
Explaining interference patterns is simple:
  • Bright fringes appear when the peaks of light waves from the two slits align perfectly, leading to constructive interference. Here, the path difference between waves is a whole number multiple of wavelengths, denoted as \( m \lambda \).
  • Dark fringes occur when the peak of one wave aligns with the trough of another, leading to destructive interference.
Understanding this helps grasp how light behaves as a wave and demonstrates the principle of superposition.
Wavelength
Wavelength is a fundamental parameter in understanding wave behavior, including light waves. It represents the distance between consecutive peaks (or troughs) of a wave. In the context of the two-slit experiment, the wavelength (\( \lambda \)) is crucial in determining where these bright and dark fringes occur on the screen.
  • In this exercise, the wavelength of the light used is 580 nm, or nanometers, which is typical for visible light.
  • The role of wavelength in the interference pattern can be seen in how it relates to the path difference needed for constructive interference: \( m \lambda \).
A longer wavelength results in a wider spacing between fringes, while a shorter wavelength leads to narrow spacing. This highlights the beauty of light's wave nature, connecting physical dimensions like wavelength to the visual phenomena we observe.
Intensity
Intensity in wave physics describes the power transferred per unit area where the wave is present. In the case of light, intensity is a measure of the brightness observed at different interference fringes. The central bright fringe in an interference pattern typically has the maximum intensity, often denoted as \( I_0 \).
However, the actual intensity at any angular position is influenced by the slit width, introducing a single-slit diffraction component to the pattern.
  • The intensity at an angular position \( \theta \) is given by the formula \( I(\theta) = I_0 \left( \frac{\sin(\beta)}{\beta} \right)^2 \).
  • Here, \( \beta = \frac{\pi a \sin \theta}{\lambda} \), where \( a \) represents the slit width.
This relationship signifies that not only does light constructively or destructively interfere based on its wavelength and the path difference but also that the slit width modifies the overall brightness of these fringes.
Angular Position
Angular position is a key concept in describing the locations of interference maxima and minima in the two-slit experiment. The angle, \( \theta \), is the angle between the central axis perpendicular to the slits and the position on the screen where specific interference maxima occur.
Using the formula for constructive interference, \( d \sin \theta = m \lambda \) (where \( d \) is the distance between slits, \( m \) is the order of maxima, and \( \lambda \) is the wavelength), we can determine these angular positions.
  • For first-order maxima (\( m=1 \)), \( \theta_1 \) is calculated, and analogously, for second-order maxima (\( m=2 \)), \( \theta_2 \).
  • These angles are small in most practical setups, leading to approximate values often expressed in degrees, such as \( 0.0627^\circ \) and \( 0.1254^\circ \).
Thus, angular position helps us translate theoretical interference into observable optical effects, pinpointing where the bright fringes will be on a screen.

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Most popular questions from this chapter

A slit 0.360 mm wide is illuminated by parallel rays of light that have a wavelength of 540 nm. The diffraction pattern is observed on a screen that is 1.20 m from the slit. The intensity at the center of the central maximum \((\theta = 0^\circ)\) is \(I_0\). (a) What is the distance on the screen from the center of the central maximum to the first minimum? (b) What is the distance on the screen from the center of the central maximum to the point where the intensity has fallen to \(I_0\)/2?

Visible light passes through a diffraction grating that has 900 slits/cm, and the interference pattern is observed on a screen that is 2.50 m from the grating. (a) Is the angular position of the first-order spectrum small enough for sin \(\theta \approx \theta\) to be a good approximation? (b) In the first- order spectrum, the maxima for two different wavelengths are separated on the screen by 3.00 mm. What is the difference in these wavelengths?

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?

Your physics study partner tells you that the width of the central bright band in a single-slit diffraction pattern is inversely proportional to the width of the slit. This means that the width of the central maximum increases when the width of the slit decreases. The claim seems counterintuitive to you, so you make measurements to test it. You shine monochromatic laser light with wavelength \(\lambda\) onto a very narrow slit of width \(a\) and measure the width \(w\) of the central maximum in the diffraction pattern that is produced on a screen 1.50 m from the slit. (By "width," you mean the distance on the screen between the two minima on either side of the central maximum.) Your measurements are given in the table. (a) If \(w\) is inversely proportional to \(a\), then the product \(aw\) is constant, independent of \(a\). For the data in the table, graph \(aw\) versus \(a\). Explain why \(aw\) is not constant for smaller values of \(a\). (b) Use your graph in part (a) to calculate the wavelength \(\lambda\) of the laser light. (c) What is the angular position of the first minimum in the diffraction pattern for (i) \(a\) = 0.78 \(\mu\)m and (ii) \(a\) = 15.60 \(\mu\)m?

Laser light of wavelength \(500.0 \mathrm{nm}\) illuminates two identical slits, producing an interference pattern on a screen \(90.0 \mathrm{~cm}\) from the slits. The bright bands are \(1.00 \mathrm{~cm}\) apart, and the third bright bands on either side of the central maximum are missing in the pattern. Find the width and the separation of the two slits.
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