Question

On a cloudy day, a raft of size 6x6 m floats on the surface of the pond that has a depth 0, 5 m and flat horizontal bottom. Determine the area of raft shadow on the bottom of the pond. Give the answer in [ m^2 ] and round it up to a whole. The refractive index of water is 4/3.

Answer 1

"On a cloudy day" means that the light is diffused: the entire sky emits light; light rays have all possible directions. Thus, we have rays with maximum incidence angle of 90°. They will form the shadow.

From the right triangle we see that

$$d=D-h\text{ tan}\alpha_2.$$

From optics, we know that

$$\frac{\text{sin}\alpha_1}{\text{sin}\alpha_2}=n,\text{ since }\alpha_1=90°,\\\space\\ \text{sin}\alpha_2=\frac{1}{n},\\\space\\ \text{tan}\alpha_2=\frac{\text{sin}\alpha_2}{\sqrt{1-\text{sin}^2\alpha_2}}=\frac{1}{\sqrt{n^2-1}},\\\space\\ d=D-\frac{h}{\sqrt{n^2-1}}.$$

The area of the shadow is

$$A=\frac{d^2}{2}=\frac{\bigg(D-\frac{0.5}{\sqrt{(n)^2-1}}\bigg)^2}{2}=\\\space\\ =\frac{\bigg(6\sqrt2-\frac{0.5}{\sqrt{(4/3)^2-1}}\bigg)^2}{2}=31\text{ m}^2.$$