First let
us find the angle of incidence. The angle between the light ray and the
surface of the water say 'θ' can be obtained from the triangle laws as
Tan(θ) = Opp. / Adj. = h / L in our case.
i.e., Tan(θ) = 1.4 / 3.2 = 0.4375 and hence , θ = 23.629o
Hence the angle of incidence will be ' i ' = (90-θ) = 66.371o
(Since the angle of incidence is that angle the light ray makes
with the perpendicular to the surface at the point of contact).
we know that the refractive index of water is 1.33 and that of air is 1.
Hence the angle of refraction can be found as
Sin(r) = Sin(i) x (μ1 / μ2) = Sin(66.371) / 1.33 = 0.9015 / 1.33 = 0.688 and hence
the angle of refracion is 'r' = ........o.
Now again from the geometry, the horizontal base of the triangle formed in the water can be found by
Tan(r) = base / depth of pool hence,
base = Tan(r) x 2.1 = ....... m.
Hence the total distance at which the light ray touches the bottom of the pool from the wall beneath the man's foot is d = L + base = 3.2 + base = ..... m.