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Diffraction occurs for all types of waves, including sound waves. High- frequency sound from a distant source with wavelength 9.00 cm passes through a slit 12.0 cm wide. A microphone is placed 8.00 m directly in front of the center of the slit, corresponding to point \(O\) in Fig. 36.5a. The microphone is then moved in a direction perpendicular to the line from the center of the slit to point \(O\). At what distances from \(O\) will the intensity detected by the microphone be zero?

Short Answer

Expert verified
The intensity detected by the microphone will be zero at approximately 6.00 m from point O.

Step by step solution

01

Identify the Problem and Given Values

This problem involves sound wave diffraction through a slit. We need to find the points where the intensity detected by a microphone is zero, which means we are looking for the locations of the dark fringes. The given values are the wavelength of the sound wave \( \lambda = 9.00 \text{ cm} = 0.09 \text{ m} \) and the width of the slit \( a = 12.0 \text{ cm} = 0.12 \text{ m} \).
02

Use the Diffraction Equation

For a single-slit diffraction, the condition for minima (zero intensity points) is given by the equation: \[ a \sin \theta = m \lambda \] where \( m \) is the order of the minimum (\( m = \pm 1, \pm 2, \pm 3, \dots \)) and \( \theta \) is the angle from the center of the slit.
03

Find the Angle for Minima

Substitute the given values into the diffraction equation to find \( \theta \):\[ 0.12 \sin \theta = m \cdot 0.09 \]\[ \sin \theta = \frac{m \cdot 0.09}{0.12} = \frac{m}{1.333} \].
04

Calculate Specific Angles for Minima

Evaluate for the first few integers of \( m \) to find valid angles:For \( m = 1 \):\[ \sin \theta = \frac{1}{1.333} \approx 0.75 \implies \theta \approx 48.59^\circ \]For \( m = 2 \):\[ \sin \theta = \frac{2}{1.333} = 1.5 \] (not possible as the sine function cannot exceed 1, only \( m = 1 \) is valid).
05

Calculate Distance from O for the First Minimum

Using \( \tan \theta \approx \sin \theta \) for small angles, the distance \( y \) from point \( O \) to the point of zero intensity is given by:\[ \tan \theta \approx \frac{y}{L} \implies y = L \tan \theta \approx L \sin \theta \]Given \( L = 8.00 \text{ m} \), and \( \sin \theta \approx 0.75 \),\[ y = 8.00 \times 0.75 = 6.00 \text{ m} \].
06

Determine Final Answer

Since only \( m = 1 \) is valid, the distance from \( O \) where the intensity is zero is approximately 6.00 m.

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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 fascinating phenomenon observed when waves encounter a narrow opening, or slit, of comparable size to their wavelength. This concept is applicable to various types of waves, including sound, light, and even water waves. When a wave passes through such a slit, it bends around the edges and spreads out. This bending causes the wave to interfere with itself, resulting in patterns of alternating light and dark bands called fringes.

The single-slit diffraction pattern is characterized by a central bright fringe, which is the most intense, followed by a series of progressively dimmer and narrower fringes on either side. The central bright fringe is due to constructive interference, where the waves overlap in phase, while the dark fringes arise from destructive interference, where they cancel each other out. The mathematical analysis of this pattern helps us understand how different parameters like slit width and wavelength influence the diffraction process.
Wavelength Calculation
Wavelength calculation is crucial when working with diffraction patterns, as it dictates how waves will diffract through a slit. Wavelength (\( \lambda \)) is defined as the distance between consecutive points of a wave that are in phase, such as crest to crest. In diffraction contexts, knowing the wavelength allows us to predict interference patterns and locate fringes accurately.

For instance, if the wavelength and the slit width are known, they can be used in formulas to determine angles and positions where destructive interference (minima) will occur. In the problem discussed, we have a sound wave with a wavelength of 9.00 cm, which was converted to meters for ease in calculations. Accurate conversion and calculation are key because they set the stage for reliable predictions of diffraction effects. Such predictions are essential in various fields, from engineering applications to the design of acoustics for spaces.
Diffraction Minima
Diffraction minima refer to the specific points in a diffraction pattern where the wave intensity is zero, due to destructive interference of waves. These spots are regions where the path difference between diffracted waves from various parts of the slit leads to them arriving out of phase and cancelling each other out. Identifying minima is essential because it marks the boundaries where light or sound appears to 'cut out'.

The equation \( a \sin \theta = m \lambda \) is used to calculate the minima in single-slit diffraction scenarios, where \(a\) is the slit width, \( \theta \) is the angle of diffraction, \( \lambda \) is the wavelength, and \( m \) is the order of the minimum. For valid minima, \( m \) must be an integer (\( \pm 1, \pm 2, \ldots \)). In our example, only the first minimum (\( m = 1 \)) was valid, resulting in calculated angle and distance from point O. It's these precise conditions that guide the placement of microphones or other detecting devices for optimal soundwave analysis.

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Most popular questions from this chapter

Monochromatic light from a distant source is incident on a slit 0.750 mm wide. On a screen 2.00 m away, the distance from the central maximum of the diffraction pattern to the first minimum is measured to be 1.35 mm. Calculate the wavelength of the light.

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?

Different isotopes of the same element emit light at slightly different wavelengths. A wavelength in the emission spectrum of a hydrogen atom is 656.45 nm; for deuterium, the corresponding wavelength is 656.27 nm. (a) What minimum number of slits is required to resolve these two wavelengths in second order? (b) If the grating has 500.00 slits/mm, find the angles and angular separation of these two wavelengths in the second order.

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.

Why is visible light, which has much longer wavelengths than x rays do, used for Bragg reflection experiments on colloidal crystals? (a) The microspheres are suspended in a liquid, and it is more difficult for x rays to penetrate liquid than it is for visible light. (b) The irregular spacing of the microspheres allows the longerwavelength visible light to produce more destructive interference than can x rays. (c) The microspheres are much larger than atoms in a crystalline solid, and in order to get interference maxima at reasonably large angles, the wavelength must be much longer than the size of the individual scatterers. (d) The microspheres are spaced more widely than atoms in a crystalline solid, and in order to get interference maxima at reasonably large angles, the wavelength must be comparable to the spacing between scattering planes.

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