Answer:
a The diagram of the situation is shown on the first uploaded image
b the angle of incidence beam striking the water is [tex]\theta = 49.63^o[/tex]
c the angle of refraction beam striking the water is [tex]r = 59.7^o[/tex]
d The angle the refracted beam make with respect to the horizontal is [tex]= 30.3^o[/tex]
e The height of the target above sea level is
[tex]h= 125.05m[/tex]
Explanation:
From the diagram we see that the angle of the beam striking the water is
[tex]tan \theta = \frac{100}{85}[/tex]
[tex]\theta = tan^{-1}(\frac{100}{85} )[/tex]
[tex]= 49.63^o[/tex]
According to Snell's law
[tex]\mu_{water} *sin(i) = \mu_{air} *sin(r)[/tex]
Where [tex]\mu_{water }[/tex] is the refractive index of water = 1.333
[tex]i[/tex] is the angle of incidence
[tex]\mu_{air}[/tex] is the refractive index of air = 1
r is the angle of refraction
Substituting values accordingly
[tex]1.33 * sin (40.37) = 1 * sin(r)[/tex]
Making r the subject of the formula
[tex]r = sin^{-1}(\frac{1.333 *sin(40.37)}{1})[/tex]
[tex]= 59.7^o[/tex]
looking at the diagram we can see that to obtain the angle the refraction beam makes with the horizontal by subtracting the angle refraction from 90°
i.e [tex]90 -59.7 = 30.3[/tex] °
From the diagram we see that the height target above sea level can be obtained by this relation
[tex]tan \theta = \frac{h}{214}\\[/tex]
Where h is the is the height
[tex]tan(30.3) = \frac{h}{214}[/tex]
[tex]h = 214 * tan (30.3)[/tex]
[tex]=125.05m[/tex]
