Question

An array of wave emitters, as shown in the figure, emits a wave of wavelength \(\lambda\) that is to be detected at a distance \(L\) directly above the rightmost emitter. The distance between adjacent wave emitters is \(d\). a) Show that when \(L \gg d\), the wave from the \(n\) th emitter (counting from right to left with \(n=0\) being the rightmost emitter) has to travel an extra distance of \(\Delta s=n^{2}\left(d^{2} / 2 L\right)\). b) If \(\lambda=d^{2} / 2 L\), will the interference at the detector be constructive or destructive? c) If \(\lambda=d^{2} / 2 L=10^{-3} \mathrm{~m}\) and \(L=1.00 \cdot 10^{3} \mathrm{~m},\) what is \(d\), the distance between adjacent emitters?

Short answer

Based on the given conditions and information, we have determined that the interference pattern generated by the array of emitters is constructive if the path difference is an integer multiple of the wavelength. The expression for the extra distance traveled by the waves emitted from each emitter is given by \(\Delta s = n^2\left(\frac{d^2}{2L}\right)\), where \(n\) must be an integer. Given the values for the wavelength (\(\lambda = 10^{-3}~\text{m}\)) and length (\(L = 1.00 \times 10^3~\text{m}\)), we calculated that the distance between adjacent emitters in the array is \(d = 1~\text{m}\).

Step-by-step solution

Part a: Determine the extra distance traveled by the wavesfrom each emitter

To determine the extra distance for the waves emitted from each emitter, we can use the Pythagorean theorem. The distance traveled by the waves from the rightmost emitter is \(L\). For any other emitter at position \(n\), we can find the distance traveled as the hypotenuse of a right triangle with legs of length \(L\) and \(nd\). $$D_n = \sqrt{L^2+(nd)^2}$$ The extra distance when compared to the rightmost emitter (where \(n=0\)) is given by \(\Delta s_n = D_n - L\). Therefore, $$\Delta s_n = \sqrt{L^2+(nd)^2} - L$$ Since \(L \gg d\), we can make the approximation \(1 \approx \sqrt{1+x} \approx 1+\frac{x}{2}\) where \(x\) is small. Therefore, $$\Delta s_n \approx L\left(\sqrt{1+\left(\frac{nd}{L}\right)^2}-1\right) \approx L\left(\frac{1}{2}\left(\frac{nd}{L}\right)^2\right) = n^2\left(\frac{d^2}{2L}\right)$$ Thus, the extra distance is given by \(\Delta s = n^2\left(\frac{d^2}{2L}\right)\), as we wanted to show.

Part b: Determine the type of interference

To determine if the interference will be constructive or destructive at the detector point, we can use the criteria for path difference. When the path difference is an integer multiple of the wavelength (i.e., \(\Delta s = m\lambda\), where \(m\) is an integer), the interference is constructive. In contrast, when the path difference is an integer multiple of the half-wavelength (i.e., \(\Delta s = (m+\frac{1}{2})\lambda\)), the interference is destructive. Given the condition \(\lambda=d^2 / 2L\), let's substitute the expression for \(\Delta s\), $$\Delta s = m\lambda \Longrightarrow n^2\left(\frac{d^2}{2L}\right) = m\left(\frac{d^2}{2L}\right)$$ From this, we can see that the interference will be constructive as long as \(n^2 = m\), meaning \(n\) is an integer.

Part c: Calculate the distance between adjacent emitters

Given the values for \(\lambda\) and \(L\), we can calculate the distance between adjacent emitters \(d\) using the formula \(\lambda=d^{2} / 2 L\): $$d^2=2L\lambda$$ Given that \(\lambda=10^{-3} \mathrm{~m}\) and \(L=1.00\times 10^3 \mathrm{~m}\), we plug the values into the formula to get $$d^2=(2)(1.00\times 10^3 \mathrm{~m})\left(10^{-3}\mathrm{~m}\right)$$ Now, we solve for \(d\): $$d=\sqrt{(2)(1000)(0.001)}\mathrm{m}=1\mathrm{~m}$$ Thus, the distance between adjacent emitters is \(d = 1 \mathrm{~m}\).

Key concepts

Constructive and Destructive Interference

Understanding wave interference is crucial for grasping various phenomena in physics, such as sound, light, and other wave types. Interference occurs when two or more waves superpose to form a resultant wave of greater, lower, or the same amplitude.

Constructive interference happens when waves meet in phase; that is, their peaks (crests) and troughs align to produce a wave with a larger amplitude. This type of interference can result in a sound that's louder or light that's brighter. On the other hand, destructive interference occurs when waves meet out of phase, meaning the peak of one wave aligns with the trough of another, cancelling each other out to produce a weaker wave or, in the case of complete cancellation, no wave at all.

To better understand this concept, imagine clapping your hands versus clapping one hand against the air next to the other hand: in the first scenario, the claps `construct' together to produce a louder sound, while in the second, there is much less sound produced due to the lack of `destructive' interference from the second hand.

Path Difference

Path difference plays a pivotal role in determining the type of interference that occurs. It is essentially the difference in the distance that two waves have traveled upon meeting. This path difference can be responsible for changing a situation from constructive interference to destructive, and vice versa.

For waves to interfere constructively, the path difference must be an integer multiple of the wavelength, denoted as \( m\lambda \), where \( m \) is an integer, also known as the order of interference. In contrast, for destructive interference, the path difference should be an integer multiple of half the wavelength, expressed as \( (m+\frac{1}{2})\lambda \). These conditions help us to predict and control interference patterns, which is particularly useful in applications like noise cancelling headphones or creating laser beams in optical technologies.

When solving problems related to path difference, one should carefully consider the geometry of wave sources and their relative positions, as these factors can significantly affect the path that waves travel before intersecting.

Pythagorean Theorem

The Pythagorean theorem is a fundamental principle in geometry that states that in a right-angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides. Mathematically, it is expressed as \( a^2 + b^2 = c^2 \), with \( c \) representing the hypotenuse.

This theorem is not just for mathematicians; it is widely used in physics for problems involving distances and waves. In the context of wave interference and path difference, the Pythagorean theorem helps us calculate the actual distance a wave travels from its source to a point of interest, which is crucial when determining path differences and the resultant interference pattern. It allows us to work with the geometry of waves emanating from different points and to make approximations, like those used in the textbook exercise, where the distance between wave emitters and the point of detection is considerably larger than the separation of emitters themselves.