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Chapter 35: Problem 22

Two slits spaced \(0.0720 \mathrm{~mm}\) apart are \(0.800 \mathrm{~m}\) from a screen. Coherent light of wavelength \(\lambda\) passes through the two slits. In their interference pattern on the screen, the distance from the center of the central maximum to the first minimum is \(3.00 \mathrm{~mm} .\) If the intensity at the peak of the central maximum is \(0.0600 \mathrm{~W} / \mathrm{m}^{2},\) what is the intensity at points on the screen that are (a) \(2.00 \mathrm{~mm}\) and (b) \(1.50 \mathrm{~mm}\) from the center of the central maximum?

Short Answer

Expert verified
Intensity at 2 mm is \(0.0386 \text{ W/m}^2\) and at 1.5 mm is \(0.0481 \text{ W/m}^2\).

Step by step solution

01

Formula for Intensity

To find the intensity at a point on the screen, we start by noting that the intensity due to interference of two coherent sources is given by the formula: \[ I = I_0 \cos^2\left(\frac{\Delta \phi}{2}\right) \]where \( I_0 \) is the peak intensity (at the central maximum), and \( \Delta \phi \) is the phase difference.
02

Determine Phase Difference

The phase difference \( \Delta \phi \) is related to the path difference, which can be expressed in terms of the distance \( y \) from the center on the screen, the slit separation \( d \), and the distance to the screen \( L \). Using geometrical approximation, \( \Delta \phi = \frac{2\pi d \sin \theta}{\lambda} \), where \(\sin \theta = \frac{y}{L}\).
03

Calculate Wavelength

First, calculate the wavelength \( \lambda \). The position of the first minimum is given by the path difference \( \frac{\lambda}{2} \). Thus, \( d \sin \theta = \frac{\lambda}{2} \), and since the small angle approximation implies \( \sin \theta \approx \frac{y_1}{L} \), we have:\[ d \frac{y_1}{L} = \frac{\lambda}{2} \]Solving for \( \lambda \), we get:\[ \lambda = \frac{2dy_1}{L} \]The known values are \( d = 0.0720 \text{ mm} = 7.20 \times 10^{-5} \text{ m} \), \( y_1 = 3.00 \text{ mm} = 3.00 \times 10^{-3} \text{ m} \), and \( L = 0.800 \text{ m} \). Substitute these to find \( \lambda \).
04

Calculate Intensity at 2 mm

Substitute \( y = 2.00 \text{ mm} = 2.00 \times 10^{-3} \text{ m} \) in the path difference formula, calculate \( \Delta \phi \), then find intensity:1. \( \sin \theta = \frac{2.00 \times 10^{-3}}{0.800} \).2. Calculate \( \Delta \phi = \frac{2\pi d \sin \theta}{\lambda} \).3. Finally, calculate \( I = I_0 \cos^2\left(\frac{\Delta \phi}{2}\right) \) to find the intensity at \( y = 2 \text{ mm} \).
05

Calculate Intensity at 1.5 mm

Substitute \( y = 1.50 \text{ mm} = 1.50 \times 10^{-3} \text{ m} \) into the same formulas:1. \( \sin \theta = \frac{1.50 \times 10^{-3}}{0.800} \).2. Calculate \( \Delta \phi = \frac{2\pi d \sin \theta}{\lambda} \).3. Finally, calculate \( I = I_0 \cos^2\left(\frac{\Delta \phi}{2}\right) \) to find the intensity at \( y = 1.5 \text{ mm} \).

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Key Concepts

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

double-slit experiment
The double-slit experiment is a fundamental demonstration of the wave nature of light. It involves passing coherent light, such as a laser, through two closely spaced slits. As the light waves emerge from the slits, they overlap and create an interference pattern on a screen placed at some distance. This pattern consists of a series of bright and dark fringes or bands.
Several factors influence this pattern:
  • Wavelength (\( \lambda \)): The distance between successive peaks of the light wave. Longer wavelengths lead to wider patterns.
  • Slit Separation (\( d \)): Larger separations result in more closely spaced fringes.
  • Distance to Screen (\( L \)): The farther the screen, the more spread out the pattern.
The central fringe is the brightest, known as the central maximum, flanked by alternating bright and dark areas. These bright spots occur where the waves from the two slits meet in phase, causing constructive interference, while the dark bands occur where destructive interference cancels the waves.
wave phase difference
The phase difference between waves is a crucial aspect of the interference pattern observed in the double-slit experiment. It refers to the difference in the position of the crests and troughs of the two waves as they meet at a point on the screen. The phase difference (\( \Delta \phi \)) determines whether the interference is constructive or destructive.
Constructive interference occurs when the phase difference is an integer multiple of \( 2\pi \). At these points, the waves are in phase, reinforcing each other to produce bright fringes. Conversely, destructive interference arises when the phase difference equals an odd multiple of \( \pi \), leading to a dark fringe as the waves are out of phase.
The phase difference can be directly related to path difference (the difference in distances traveled by two waves), slit separation, and the angle (\( \theta \)) to the point on the screen:
  • Path Difference (\( \Delta x \)): Given by \( d \sin \theta \).
  • Relationship to Phase Difference: \( \Delta \phi = \frac{2\pi \Delta x}{\lambda} \).
This gives us a practical way to calculate phase differences for specific positions on the interference pattern.
intensity calculation
Calculating the intensity of the interference pattern at any point on the screen involves understanding how the superimposed waves interact. The intensity (\( I \)) at a point is dependent on the square of the amplitude of the resultant wave formed by the overlap of individual waves.
From the step-by-step solution, the intensity formula is given by:\[ I = I_0 \cos^2\left(\frac{\Delta \phi}{2}\right) \]Where:
  • \( I_0 \) is the peak intensity at the central maximum.
  • \( \Delta \phi \) is the phase difference.
The formula shows that the intensity varies with the cosine squared of half the phase difference. Thus, wherever \( \Delta \phi \) leads the cosine squared function to peak, you find maxima (bright fringes), whereas minima (dark fringes) occur at zeros of the cosine function.
To find the intensity at specific points, such as 2.00 mm and 1.50 mm from the center, it is essential to calculate \( \Delta \phi \) using the given slit separation and screen distance, then use the intensity formula. This method reveals how the overlap of waves creates a varying pattern of light on the screen.

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Most popular questions from this chapter

Two rectangular pieces of plane glass are laid one upon the other on a table. A thin strip of paper is placed between them at one edge so that a very thin wedge of air is formed. The plates are illuminated at normal incidence by 546-nm light from a mercury-vapor lamp. Interference fringes are formed, with 15.0 fringes per centimeter. Find the angle of the wedge

A uniform film of TiO\(_2\) , 1036 nm thick and having index of refraction 2.62, is spread uniformly over the surface of crown glass of refractive index 1.52. Light of wavelength 520.0 nm falls at normal incidence onto the film from air. You want to increase the thickness of this film so that the reflected light cancels. (a) What is the \(minimum\) thickness of TiO\(_2\) that you must \(add\) so the reflected light cancels as desired? (b) After you make the adjustment in part (a), what is the path difference between the light reflected off the top of the film and the light that cancels it after traveling through the film? Express your answer in (i) nanometers and (ii) wavelengths of the light in the TiO\(_2\) film.

Coherent light that contains two wavelengths, 660 nm (red) and 470 nm (blue), passes through two narrow slits that are separated by 0.300 mm. Their interference pattern is observed on a screen 4.00 m from the slits. What is the distance on the screen between the first-order bright fringes for the two wavelengths?

Two light sources can be adjusted to emit monochromatic light of any visible wavelength. The two sources are coherent, 2.04 \(\mu\)m apart, and in line with an observer, so that one source is 2.04 \(\mu\)m farther from the observer than the other. (a) For what visible wavelengths (380 to 750 nm) will the observer see the brightest light, owing to constructive interference? (b) How would your answers to part (a) be affected if the two sources were not in line with the observer, but were still arranged so that one source is 2.04 \(\mu\)m farther away from the observer than the other? (c) For what visible wavelengths will there be \(destructive\) interference at the location of the observer?

Two slits spaced 0.450 mm apart are placed 75.0 cm from a screen. What is the distance between the second and third dark lines of the interference pattern on the screen when the slits are illuminated with coherent light with a wavelength of 500 nm?
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