Coherent, monochromatic light of wavelength \(450.0 \mathrm{nm}\) is emitted from two locations and detected at another location. The path difference between the two routes taken by the light is \(20.25 \mathrm{~cm}\). Will the two light waves interfere destructively or constructively at the detection point?

Firstly, we need to convert the given wavelength from nanometers (nm) to centimeters (cm) to match the unit of the path difference. One nanometer is equal to \(1\times 10^{-7}\,\mathrm{cm}\). Therefore, the wavelength (λ) in centimeters can be calculated as: λ = \((450.0 \times 10^{-7})\,\mathrm{cm} = 4.5 \times 10^{-5}\,\mathrm{cm}\)

Next, we need to find the phase difference between the two light waves to determine the type of interference. The phase difference (Δφ) can be calculated by dividing the path difference (Δd) by the wavelength (λ) and then multiplying the result by \(2π\): Δφ = \(\frac{\Delta d}{\lambda} \times 2\pi\) Here, the path difference (Δd) is given as \(20.25\,\mathrm{cm}\). Now put the values into the formula: Δφ = \(\frac{20.25\,\mathrm{cm}}{4.5 \times 10^{-5} \,\mathrm{cm}} \times 2\pi = 1800\pi\)

Finally, now we can analyze the type of interference based on the phase difference. There are two conditions: 1. Constructive Interference: When the phase difference is an integer multiple of \(2\pi\), constructive interference occurs. In this case, Δφ = \(2n\pi\), where n is an integer (0, 1, 2, ...). 2. Destructive Interference: When the phase difference is an odd integer multiple of \(\pi\), destructive interference occurs. In this case, Δφ = \((2n+1)\pi\), where n is an integer (0, 1, 2, ...). Now let's analyze the calculated phase difference: Δφ = \(1800\pi = 2(900)\pi\) Here, Δφ is an even integer multiple of \(2\pi\), which means the two light waves interfere constructively at the detection point.