Question

Two light sources can be adjusted to emit monochromatic light of any visible wavelength. The two sources are coherent, 2.04 μm apart, and in line with an observer, so that one source is 2.04 μm farther from the observer than the other.

For what visible wavelengths (400 to 700 nm) will the observer see the brightest light, owing to constructive interference?

Answer 1

The question deals with finding visible wavelengths that result in constructive interference for two coherent light sources 2.04 μm apart. By applying the constructive interference condition (mλ = d) and calculating for the visible spectrum, we can identify the wavelengths where the observer sees the brightest light.

Step-by-step explanation:

The student's question pertains to constructive interference in the context of wave optics, more specifically determining the wavelengths of visible light that will result in constructive interference for two coherent light sources separated by a distance of 2.04 μm.

Constructive interference occurs when two waves combine to produce a resultant wave with a greater amplitude. For constructive interference to occur, the path difference between the two light waves must be an integer multiple of the wavelength, satisfying the condition:

mλ = d

where m is an integer (the order of interference), λ is the wavelength, and d is the path difference between the two sources. Since the sources are separated by 2.04 μm and one source is 2.04 μm farther from the observer, the path difference is 2.04 μm.

To find the wavelengths for constructive interference within the visible spectrum (400 nm to 700 nm), we solve for λ:

λ = d/m

We calculate the values of λ for different orders of m that fall within the visible range. Wavelengths that satisfy this condition and are within the visible spectrum will be the ones where the observer sees the brightest light due to constructive interference.