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 µm.
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 µm.
Answer:
λ = 428.6 nm
Explanation:
Hello,
In this case, we must remember that the Young’s double slit experiment is described by the expression
:
d sin θ = m λ
For constructive interference
, and:
d sin θ = (m + ½) λ
For destructive interference
, whereas d accounts for the distance between the slits, λ for the wavelength and m for an integer that describes the order of interference
. Thus, for the given angle 30º, the distance between the slits is 3.00 μm or 3.00 10⁻⁶ m and the order of interference is 3; we therefore use the destructive interference equation to compute the wavelength as shown below:
λ = 3×10⁻⁶ sin (30) / (3 +1/2)
λ = 4.286 10⁻⁷ m
Or in manometers:
λ = 428.6 nm
Best regards.