The width of the central maxima, formed from light of wavelength 575 nm behind a single slit that has a width of 115 μm, is 1.15 cm, what is the distance between the slit and the screen?

 L  = 1.15 m

Explanation:

The diffraction phenomenon is described by the equation

        a sin θ = m λ

Where a is the width of the slit, λ  the wavelength and m is an integer, the order of diffraction is left.

The diffraction measurements are made on a screen that is far from the slit, and the angles in the experiment are very small, let's use trigonometry

          tan θ = y / L

          tan θ = sint θ / cos θ≈ sin θ

We substitute in the first equation

           a (y / L) = m λ

The first maximum occurs for m = 1

The distance is measured from the center point of maximum, which coincides with the center of the slit, in this case the distance is the total width of the central maximum, so the distance (y) measured from the center is

         y = 1.15 / 2 = 0.575 cm

         y = 0.575 10⁻² m

Let's clear the distance to the screen (L)

       L = a y / λ  

Let's calculate

     L = 115 10⁻⁶  0.575 10⁻² / 575 10⁻⁹

     L  = 1.15 m

the second one and the forth one are in the purple because they are opinions and the rest are in the yellow

explanation: