4.) In the following figure a 2,5 m-long vertical pole extends from the bottom of a
swimming pool to a point 30cm above the water. Sunlight is incident at angle
θ=50o
. What
is the length of the shadow of the pole on the level bottom of the pool?
Solution:
Data:
Refractive index of water:
n2=1,33
d2=h2∗tan (θr)
As we know that the angle of incidence is 50 ° we search the angle of refraction:
Sen
(
θr
)
=n1
n2
∗Sen(θi)
Sen
(
θ
r
)
=1
1,33∗Sen (50)
Sen
(
θ
r
)
=0,57
(
θ
r
)
=Sen
−1
(
0,57
)
(
θ
r
)
=34,75 °
Now replacement in the initial formula:
d
2
=130 cm∗tan (34,75)
