Question

Coherent, monochromatic light of wavelength \(450.0 \mathrm{nm}\) is emitted from two locations and detected at another location. The path length 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 interference at the detection point will be destructive.

Step-by-step solution

Find the path length difference in meters

To simplify our calculations, we must first convert the path length difference into meters. Given path length difference = \(20.25 cm\) Now convert it into meters: \(20.25 cm \times \frac{1 m}{100 cm} = 0.2025 m\)

Calculate the phase difference

Now we have to determine the phase difference between the two light waves, which can be found using the following formula: \(\Delta \phi = \frac{2 \pi}{\lambda} \times \Delta L\) Where \(\Delta \phi\) is the phase difference, \(\lambda\) is the wavelength, and \(\Delta L\) is the path length difference. Given the wavelength, \(\lambda = 450.0 nm = 450.0 \times 10^{-9} m\), we can now plug in the values into the formula: \(\Delta \phi = \frac{2 \pi}{450.0 \times 10^{-9} m} \times 0.2025 m\) Now, calculate the phase difference: \(\Delta \phi = \frac{2 \pi}{450.0 \times 10^{-9} m} \times 0.2025 m \approx 2.849 \times 10^3 \ rad\)

Determine the type of interference

We will now divide the phase difference by \(\pi\) to find out whether the fractional part is close to an integer or an odd integer: \(\frac{\Delta \phi}{\pi} \approx \frac{2.849 \times 10^3}{3.14} \approx 907.64\) Since 907.64 is very close to the odd integer 907, we can conclude that the phase difference is approximately an odd-integer multiple of \(\pi\).

Conclusion

As the phase difference between the two light waves is approximately an odd-integer multiple of \(\pi\), we can conclude that the interference at the detection point will be destructive.