Question

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?

Short answer

Answer: The two light waves interfere constructively at the detection point.

Step-by-step solution

Wavelength Conversion

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 = \((450.0 \times 10^{-7})\,\mathrm{cm} = 4.5 \times 10^{-5}\,\mathrm{cm}\)

Determine the Phase Difference

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 = \(\frac{20.25\,\mathrm{cm}}{4.5 \times 10^{-5} \,\mathrm{cm}} \times 2\pi = 1800\pi\)

Check the Conditions for Constructive and Destructive Interference

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.

Key concepts

Wavelength Conversion

Whenever we talk about the measurement of light waves, it's crucial to align the units for accurate calculations. In the exercise, the original wavelength of light was given as 450 nm, which stands for nanometers. Since the path difference in the problem was measured in centimeters, converting the wavelength to centimeters was a necessary step.

This conversion involves multiplying the wavelength in nanometers by the conversion factor from nanometers to centimeters, which is 1 nm = \(1 \times 10^{-7}\) cm. Therefore, the conversion looks like this:

\(450.0 \times 10^{-7} \mathrm{cm}\) or simplified to \(4.5 \times 10^{-5}\) cm.

By converting units early, you ensure that all measurements are comparable, making it easier to compute values accurately in the following steps.

Phase Difference

Understanding the phase difference helps in predicting how waves interact with each other. In simple terms, phase difference is the measure of how "out of step" the waves are with one another.

To find this, we use the formula:

• \(\Delta \phi = \frac{\Delta d}{\lambda} \times 2\pi \)

where \(\Delta d\) is the path difference, and \(\lambda\) is the wavelength that we converted in the previous step. Plugging in the values, the path difference is 20.25 cm, and the wavelength is \(4.5 \times 10^{-5}\) cm.

After inserting these values into the formula, we find that \(\Delta \phi = 1800\pi\).

This indicates how much one wave leads or lags behind the other. Knowing the phase difference is crucial for determining the type of interference that will occur at the detection point.

Constructive and Destructive Interference

Light interference occurs when waves overlap, and the resulting wave amplitude can indicate constructive or destructive interference. This is central to understanding various optical phenomena. Let's break down what each type means:

- **Constructive Interference:** This takes place when two waves arrive in phase, meaning their peaks align. The phase difference between them is an integer multiple of \(2\pi\). Mathematically expressed as \(\Delta \phi = 2n\pi\), where \(n\) is any integer.

In this condition, the wave amplitudes add together, resulting in a brighter and more intense wave. In our exercise, \(\Delta \phi = 1800\pi\), which simplifies to \(2(900)\pi\). Clearly an integer multiple of \(2\pi\), indicating constructive interference.

- **Destructive Interference:** This occurs when waves arrive out of phase, meaning peaks and troughs align with each other. The phase difference fits the formula \(\Delta \phi = (2n+1)\pi\), again with \(n\) as an integer.

Here, wave amplitudes cancel out, leading to reduced or nullified detected light.-

The exercise shows constructive interference, meaning the light waves reinforce each other, rather than cancel out.