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A source \(S\) and detector \(D\) are \(d\) apart. The direct wave from \(S\) is in phase with the wave from \(S\) rellccted from a horizontal layer at altitude \(H .\) When the layer rises by \(h\) no signal is detected. Find the wavelength of the signal from the source.

Short Answer

Expert verified
The wavelength of the signal is \( \lambda = 4h \).

Step by step solution

01

Understanding the Problem Statement

We have a source and a detector separated by a distance \( d \). Two types of waves are considered – a direct wave from source \( S \) and a reflected wave from a horizontal layer at altitude \( H \). These waves are initially in phase. With an increase in the layer's height by \( h \), no signal is detected implying a phase difference of \( \pi \) or an equivalent half-wavelength path difference.
02

Calculate Path Difference

The path difference between the direct wave path and the reflected wave path before height increase is zero since they are in phase. Upon the increase in height \( h \), the path length of the reflected wave increases by \( 2h \) (due to the round trip). The condition of no signal indicates a path difference of half a wavelength, \( 2h = \frac{\lambda}{2} \).
03

Derive the Wavelength Formula

Given the path increase condition, \( 2h = \frac{\lambda}{2} \), solve for the wavelength \( \lambda \). Rearranging the equation gives \( 2h \cdot 2 = \lambda \), thus \( \lambda = 4h \).
04

Concluding the Solution

The calculation has shown the wavelength \( \lambda \) is equal to \( 4h \). Therefore, the wavelength of the signal from the source is given by \( \lambda = 4h \).

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Key Concepts

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

Path Difference
In wave interference, path difference is crucial for determining how two waves interact. When waves travel from a source to a detector, the difference in the distance traveled by each wave can cause them to interfere either constructively or destructively. In the exercise, there's a direct wave and a wave that gets reflected off a horizontal layer. Initially, these waves are perfectly in phase, so their path difference is zero. This means they are reinforcing each other, leading to detectable signals.

However, when the layer's height increases by a small amount, the path length of the reflected wave also increases. This change in path length is described as a round trip, meaning it doubles the height increase (i.e., the path difference becomes 2h). When the waves no longer constructively interfere, it indicates the path difference has changed to a half-wavelength, causing destructive interference, hence when no signal is detected.
Wavelength
The wavelength is a fundamental aspect of understanding wave behavior. It is the distance between two consecutive corresponding points on a wave, such as from peak to peak. In the problem, the wavelength of the signal from the source affects how the waves interact after being reflected.

By observing that no signal is detected when the height of the reflective layer increases by h, we figure out that the path difference equals half of the wavelength (i.e., causing phase difference). Using this information, and with the condition that the path difference increased by 2h, we derive the wavelength formula:
  • \[ \lambda = 4h \]
This means each full wavelength equals four times the height increase of the reflective layer.
Phase Difference
Phase difference refers to the change in phase between two waves that can cause constructive or destructive interference. When two waves meet, their phase relationship determines whether they amplify or cancel each other. In the exercise, initially both the direct wave and the reflected wave are in phase, so they interfere constructively. This means their phase difference is zero, resulting in the detector receiving a strong signal.

When the layer's height increases, the reflected wave travels further. This extra distance creates a phase difference of \(\pi\) (180 degrees), turning what was initially constructive interference into destructive interference. As a result, the waves cancel each other out, leading to no detectable signal. Understanding phase difference is key in predicting and explaining wave behavior.
Reflection
Reflection of waves involves the bouncing back of waves when they encounter a surface or medium. This reflection can significantly alter the path and phase of waves, as seen when waves hit horizontal layers. In the scenario provided, the layer acts like a mirror, allowing the incident wave from the source to reflect and travel to the detector. Initially, this reflected wave matches the direct wave in phase, creating constructive interference and a strong signal.

However, once the layer's height changes, the reflection causes the wave to travel a longer path, introducing a path difference and subsequent phase difference. The reflection here demonstrates that even small changes in the reflecting surface's position can have significant effects on wave interference patterns observed at the detector. This concept is vital in fields like acoustics and optics, where wave behavior control is important.

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Most popular questions from this chapter

Two pendulums have time-periods \(T\) and \(5 T / 4\). They start SIIM at the same time from the mean position. What will be the phase difference between them when the smaller pendulum has completed one oscillation? (a) \(60^{\circ}\) (b) \(72^{\circ}\) (c) \(90^{\circ}\) (d) \(120^{\circ}\)

The speed of a transverse wave, going on a wire having a length \(50 \mathrm{~cm}\) and mass \(5.0 \mathrm{~g}\), is \(80 \mathrm{~m} / \mathrm{s}\). The area of cross-section of the wire is \(1.0 \mathrm{~mm}^{2}\) and its Young's modulus is \(16 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2}\). Find the extension of the wire over its natural length.

\(\Lambda\) wave represented by \(y=2 \cos (4 x-\pi t)\) is supcrposed with another wave to form a stationary wave such that the point \(x=0\) is a node. The equation of other wave is (a) \(2 \sin (4 x+\pi t)\) (b) \(-2 \cos (4 x-\pi t)\) (c) \(-2 \cos (4 x+\pi t)\) (d) \(-2 \sin (4 x-\pi t)\)

(a) Let \(L \rightarrow\) stationary, \(\nu=580 \mathrm{II} \mathrm{z}\) \(s \rightarrow\) moving with \(40 \mathrm{~km} / \mathrm{hr}\) and aproaching \(w \rightarrow 40 \mathrm{~km} / \mathrm{hr}\) and supporting \(v=1200 \mathrm{~km} / \mathrm{hr}\) So \(v^{\prime}=\left[\frac{v+\omega}{(v+\omega)-v_{s}}\right] v\) \(\Rightarrow \quad v^{\prime}=\frac{1240}{1200} \times 580=\frac{1798}{3}=599.33 \mathrm{IIz}\) (b) At a distance \(1 \mathrm{~km}\) before the train whistles and let the driver heard the echo at time \(l=t^{\prime}\). then at this time, total distance travelled by wave and train \(=2 \mathrm{~km}\) Now time after which first wave reach to the hill in time interval \(t_{1}=\frac{1}{1200} \mathrm{hr}\) after reflecting at \(t=t^{\prime}\), echo will heard, then distance travellcd by it is \(x_{1}=\left(t^{\prime}-\frac{1}{1200}\right) \times 1200\) and distance travelled by the train is \(x_{2}=\left(t^{\prime}-\frac{1}{1200}\right) 40\) As \(x_{1}+x_{2}=1 \mathrm{~km}\) \(\therefore\left(t^{\prime}-\frac{1}{1200}\right) \times 12040+\left(t^{\prime}-\frac{1}{1200}\right) 40\) \(1240 t^{\prime}=\frac{1}{30} \quad \therefore \quad t^{\prime}=\frac{61}{30 \times 1240}\) and required distance, $$ x_{1}=\left(t^{\prime}-\frac{1}{1200}\right) \times 1200 $$$=\left(\frac{61}{30 \times 1240}-\frac{1}{1200}\right) \times 1200=\frac{61 \times 1200}{30 \times 1240}-1\( \)=\frac{30}{31} \times 1000 \mathrm{~m}=967.74 \mathrm{~m}\( Frequency of sound heard by the driver, \)v^{\prime \prime}=\left(\frac{(v-\omega)+v_{2}}{v-\omega}\right) v\( \)=\left(\frac{1200-40+40}{1200-40}\right) \times 599.33\( \)\therefore \quad v^{\prime \prime}=\frac{1200}{1160} \times 599.33=620 \mathrm{~Hz}$

A string of length \(L\) is stretched along the \(x\)-axis and is rigidly clamped at its two ends. It undergoes transverse vibrations. If \(n\) is an integer, which of the followings relations may represent the shape of the string at any time \(t ?\) (a) \(y-\Lambda \sin \left(\frac{n \pi x}{L}\right) \cos \omega\) (b) \(y-\Lambda \sin \left(\frac{n \pi x}{L}\right) \sin \alpha\) (c) \(y-A \cos \left(\frac{n \pi x}{L}\right) \cos \omega\) (d) \(y-A \cos \left(\frac{n \pi x}{L}\right) \sin \omega\)

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