Question

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

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

Step-by-step solution

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 \( a = 12.0 \text{ cm} = 0.12 \text{ m} \).

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.

Find the Angle for Minima

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

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).

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 \( \sin \theta \approx 0.75 \),\[ y = 8.00 \times 0.75 = 6.00 \text{ m} \].

Determine Final Answer

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

Key concepts

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.