It's nighttime, and you've dropped your goggles into a 3.2-mm-deep swimming pool. If you hold a laser pointer 1.0 mm above the edge of the pool, you can illuminate the goggles if the laser beam enters the water 1.9 mm from the edge. How far are the goggles from the edge of the pool?

The distance of the google to the edge of the pool is 4.75 mm

Explanation:

The angle made by the laser and surface of the pool denoted as β, is calculated as follows;

The incident of the laser beam = 90 - 27.74 = 62.26°

Apply Sneil's law to calculate refracted angle of air-water interface

Refractive index of air, na = 1

Refractive index of water, nw = 1.33

na(sinθi) = nw(sinθr)

where;

θi is the incident of the laser beam

θr is refracted angle of the laser beam in water

The displacement of the refracted laser beam, d is calculated as follows;

The distance of the google to the edge of the pool = 1.9 mm + 2.85 mm

= 4.75 mm

Check the image uploaded for the diagram and for better understanding.

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 .