Answer:
The distance of the google to the edge of the pool is 4.75 mm
Step-by-step explanation:
The angle made by the laser and surface of the pool denoted as β, is calculated as follows;
![tan \beta = (1)/(1.9) = 0.526\\\\\beta =tan^(-1)(0.526) =27.74^o](https://img.qammunity.org/2021/formulas/physics/college/cj1yjd47gnki7ezr5w42we9i7u6acg92rv.png)
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
![sin \theta_r = (n_a(sin \theta_i))/(n_w) = (1(sin62.26))/(1.33) =0.666\\\\\theta_r =sin^(-1)(0.666) = 41.72^o](https://img.qammunity.org/2021/formulas/physics/college/qm9i1uqurxg2sox3kzqcugvka8zr97zjmf.png)
The displacement of the refracted laser beam, d is calculated as follows;
![tan \theta_r = (d)/(3.2) \\\\d = 3.2*tan(41.72) = 2.85 \ mm](https://img.qammunity.org/2021/formulas/physics/college/tioadgfbbplhczdyjoxjasrk9ncp51gzvs.png)
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.