Answer:
the googles are 5.3 m from the edge
Explanation:
Given that
depth of pool , d = 3.2 m
Now, let i be the angle of incidence
a laser pointer 0.90 m above the edge of the pool and laser beam enters the water 2.2 m from the edge
⇒tan i = 2.2/0.9
[tex]i=arctan(2.2/.90)[/tex]
solving we get
i = 67.8°
Using snell's law ,
n1 ×sin(i) = n2 ×sin(r)
n1= refractive index of 1st medium= 1
n2= refractive index of 2nd medium = 1.33
r= angle of reflection
therefore,
[tex]1\times sin(67.8) = 1.33\times sin(r)[/tex]
r = 44.1°
Now,
distance of googles = 2.2 + d×tan(r)
distance of googles = 2.2 + 3.2×tan(44.1)
distance of googles = 5.3 m
the googles are 5.3 m from the edge