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.

Answers

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:

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 λFor constructive interference

, and:

d sin θ = (m + ½) λFor 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 nmBest regards.