Calculate the wavelength of light that has its third minimum at an angle of 30.0º when falling on double slits separated by 3.00~/mu/text{m}3.00 μm.

The condition for a minimum of the intensity of light falling on double slits is given by:

d · sinα = (n + 1/2) λ

Therefore, we can solve for λ:

λ = (d · sinα) / (n + 1/2)

= (3 · sin30) / (3 + 1/2)

= 0.429 μm

Therefore, the light has a wavelength of 429nm.