A double slit experiment is performed with monochromatic light. The separation between the slits is 1.150mm apart and an interference pattern is noticed on a screen 3.90m away. If the second fringe is 4.20mm from the central fringe, what wavelength and colour is the light? (You need to google wavelengths

Isn't this a standard formula? What is it about this you do not understand?

x=kLλ/d

λ=xd/kL=
=4.2•10⁻³•1.15•10⁻³/3.9•2=6.2•10⁻⁷ m
red

To find the wavelength and color of the light in the given double slit experiment, we can use the formula for the fringe spacing in a double-slit interference pattern:

Δy = (λ * L) / d

Where:
Δy is the distance between fringes
λ is the wavelength of the light
L is the distance from the double slit to the screen
d is the separation between the slits

Given information:
Δy = 4.20 mm (0.00420 m)
L = 3.90 m
d = 1.150 mm (0.00115 m)

Rearranging the formula, we can solve for the wavelength (λ):

λ = (Δy * d) / L

Now we can substitute the values to calculate the wavelength:

λ = (0.00420 m * 0.00115 m) / 3.90 m

λ ≈ 1.241 * 10^-6 m

To determine the color of the light, we can use the relationship between wavelength and color. The visible light spectrum ranges from approximately 400 nm (violet) to 700 nm (red). Converting the wavelength to nanometers:

λ ≈ 1.241 * 10^-6 m * 10^9 nm/m

λ ≈ 1241 nm

The calculated wavelength of approximately 1241 nm corresponds to the infrared region of the electromagnetic spectrum. Therefore, the color of the light is outside the visible range and not perceivable to the human eye.