Final answer:
The maximum wavelength for destructive interference where the path difference is 25.3 m is found by increasing an integer m until the wavelength equation 2 * 25.3 m / (m + 0.5) gives a value just under 25.3 m. The correct answer is 2.40 m.
Step-by-step explanation:
To find the maximum wavelength for destructive interference at the given location from two coherent, in-phase wave sources, we can use the concept that destructive interference occurs when the path difference between the two waves is an odd multiple of half the wavelength. It means that the path length difference must be (m + 0.5) λ, where m is an integer, and λ is the wavelength.
Given that the location is 30 m from source A and 4.70 m from source B, the path length difference is 30 - 4.70 = 25.3 m. For destructive interference, this path difference should be equal to (m + 0.5) λ. We look for the largest possible value of λ which still satisfies an integer m.
The maximum wavelength is therefore 2 * 25.3 m / (m + 0.5), where m is the largest integer that does not lead to a λ greater than 25.3 m. Since we are looking for the maximum λ, we start with m = 0, which gives a wavelength of 50.6 m, which is not permissible here (greater than 25.3 m). Next, m = 1 gives λ = 2 * 25.3 / 1.5 = 33.7333... m, which is still too high. When m = 2, we get λ = 2 * 25.3 / 2.5 = 20.24 m. Still too large, but getting closer. We keep increasing m until we find the maximum wavelength that's less than 25.3 m. When m = 20, λ = 2 * 25.3 / 20.5 = 2.46829... m which is > 2.40 m, so the next interval (m = 21) brings us to λ = 2 * 25.3 / 21.5 = 2.35116... m, which is just a little above 2.40 m and thus not permissible.
Therefore, the correct option that represents the maximum permissible wavelength is option c. 2.40 m, which is slightly less than what would be obtained with m = 21 but more than the value from m = 20.