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.

speed of light in vacuum is 300,000 km/s

speed of light in other mediums is given by the expression v = c/n, where c is the speed of light in vacuum and n is the refractive index of medium.

for the first medium v = 200,000 km/s

  so, refractive index of medium = 300000/200000 = 1.5

    1.5 refractive index is usually for glass. so first medium is glass.

for the second medium v = 250,00 km/s

so, refractive index of medium = 300000/250000 = 1.2

    1.2 refractive index is usually for water. so second medium is water.

for the third medium v = 0 km/s

so, refractive index of medium = 300000/0 = ∞

    ∞ refractive index is usually for opaque mediums. so third medium is opaque medium.