Final answer:
The wavelength of light falling on double slits separated by 2.00 µm if the third-order maximum is at an angle of 60.0° is 1.15 µm.
Step-by-step explanation:
To find the wavelength of light falling on double slits separated by 2.00 µm with the third-order maximum at an angle of 60.0°, we utilize the formula for constructive interference in a double-slit experiment:
d sin(θ) = mλ, where:
- d is the distance between the slits,
- θ is the angle of the maximum,
- m is the order of the maximum (the integer)
- λ is the wavelength.
The given values are:
- d = 2.00 µm,
- θ = 60.0°,
- m = 3 (for third-order maximum).
Plugging these values into the formula gives us:
2.00 µm × sin(60.0°) = 3λ
Calculate sin(60.0°), which is √3/2:
2.00 µm × (√3/2) = 3λ
λ = (2.00 µm × √3/2) / 3
λ = 1.15 µm
Thus, the wavelength of the light is 1.15 µm, corresponding to option b).