Answer:
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;
[tex]tan \beta = \frac{1}{1.9} = 0.526\\\\\beta =tan^{-1}(0.526) =27.74^o[/tex]
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
[tex]sin \theta_r = \frac{n_a(sin \theta_i)}{n_w} = \frac{1(sin62.26)}{1.33} =0.666\\\\\theta_r =sin^{-1}(0.666) = 41.72^o[/tex]
The displacement of the refracted laser beam, d is calculated as follows;
[tex]tan \theta_r = \frac{d}{3.2} \\\\d = 3.2*tan(41.72) = 2.85 \ mm[/tex]
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.
