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.

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 = 3x10⁻⁶ sin (30) / (3 +1/2)

λ = 4.286 10⁻⁷ m

Or in manometers:

λ = 428.6 nm

Best regards.

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j