The observer sees the brightest light (constructive interference) at an infrared wavelength of 2.04 μm. Constructive interference doesn't occur for visible wavelengths (400 to 700 nm) in the given double-slit setup.
To determine the conditions for constructive interference in a double-slit interference pattern, we use the formula "d sin(theta) = m lambda," where:
"d" is the separation between the slits (2.04 μm),
"theta" is the angle between the central maximum and the observed maximum,
"m" is an integer representing the order of the maximum,
"lambda" is the wavelength of light.
For the observer to see the brightest light (constructive interference) at the central maximum (m = 0), "theta" must be 0, and sin(theta) = 0, implying "theta" = 0.
Therefore, the condition "d sin(theta) = m lambda" simplifies to "d = m lambda."
Now, considering "m = 1," the observer will see the brightest light when "2.04 μm = 1 x lambda." Solving for "lambda":
lambda = 2.04 μm.
This corresponds to an infrared wavelength.
For visible wavelengths (400 to 700 nm), we need to find the values of "m" for which "d sin(theta) = m lambda" is satisfied. However, within the visible spectrum, there are no integer values of "m" that fulfill this condition. Therefore, there is no constructive interference for visible wavelengths in the given setup.
The question probable may be:
Two light sources can be adjusted to emit monochromatic light of any visible wavelength. The two sources are coherent, 2.04μm apart, and in line with an observer, so that one source is 2.04μm 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?