Between this and the next assignment, we want to get a better under- standing of how light interacts with the eye. Here are two questions to get us started, focused on diffraction (i. e., the spreading of light when it passes through a narrow opening). A. To regulate the intensity of light reaching our retinas, our pupils1 change diameter anywhere from 2 mm in bright light to 8 mm in dim light. Find the angular resolution of the eye for 550 nm wavelength light at those extremes. In which light can you see more sharply, dim or bright

θ₁ = 3.35 10⁻⁴ rad ,  θ₂ = 8.39 10⁻⁵ rad

Explanation:

This is a diffraction problem for a slit that is described by the expression

       sin θ = m λ

the resolution is obtained from the angle between the central maximum and the first minimum corresponding to m = 1

      sin θ = λ / a

as in these experiments the angle is very small we can approximate the sine to its angle

        θ = λ / a

In this case, the circular openings are explicit, so the system must be solved in polar coordinates, which introduces a numerical constant.

       θ = 1.22 λ / D

where D is the diameter of the opening

 let's apply this expression to our case

indicates that the wavelength is λ = 550 nm = 550 10⁻⁹ m

the case of a lot of light D = 2 mm = 2 10⁻³ m

       θ₁ = 1.22 550 10-9 / 2 10⁻³

       θ₁ = 3.35 10⁻⁴ rad

For the low light case D = 8 mm = 8 10⁻³

      θ₂ = 1.22 550 10-9 / 8 10⁻³

      θ₂ = 8.39 10⁻⁵ rad

r=rho l/a where a is the area, l is the length and rho is resistivity so the greater the area the lesser the resistance thats why the earth pin is thicker bcos current follow the shortest path, so less resistive .

d eras and periods