Question

For the following questions, assume the wavelengths of visible light range from 380 nm to 760 nm in a vacuum. (a) What is the smallest separation (in nm) between two slits that will produce a ninth-order maximum for any visible light

Answer 1

The smallest separation between two slits that will produce a second-order maximum for any visible light is 1520 nm.

Step-by-step explanation:

In order to determine the smallest separation between two slits that will produce a second-order maximum for any visible light, we need to use the formula for the position of the bright fringes in a double-slit interference pattern:

d * sin(θ) = m * λ

where d is the slit separation, θ is the angle of the bright fringe, m is the order of the bright fringe, and λ is the wavelength of light.

For a second-order maximum (m = 2) and any visible light, we can take the largest wavelength in the visible light range (760 nm) and use it to find the smallest separation:

d = (2 * λ) / sin(θ)

d = (2 * 760 nm) / sin(90°)

d = 1520 nm

Therefore, the smallest separation between two slits that will produce a second-order maximum for any visible light is 1520 nm.

Answer 2

Step-by-step explanation:

This is an interference exercise, which the case of constructive interference is described by the expression

d sin θ = m λ

in this case they indicate that we are in the ninth order (m = 9).

To be able to observe the pattern, the dispersion angle must be less than 90º

we substitute

sin 90 = 1

d = m lang

let's calculate

d = 9 λ

d = 9 380 10⁻⁰

d = 3.42 10⁻⁶

d2 = 9 760 10⁻⁹

d2 = 6.84 10₋⁶