what is the answer to this physics problem?

[Help Me]

A boy standing in a shallow pool of water notices a penny on the bottom of the pool.

Light from the penny is coming to his eyes at an angle of <53.06> degrees from the water’s surface.

how to solve it with the answer would be appreciated. thanks!

The penny and the boy’s eyes are separated by a horizontal distance of <11.848481776731> cm and a vertical distance of <116.35198488189> cm.

How deep (in centimeters) is the water in this pool?

[anil bakshi]

E
*eye ....... |
................ |
................ |
N ------------ Z --------- refraction
................ |..... *penny (apparant)
................ | h
................ |.... O penny (real)
T <.. x1..><x2>
----------------------
snell's law of refraction>
sin i / sin r = u(water)/u(air) = 1.333
given > angle EZN = 53.06
i = incident angle = 90 - 53.06 = 36.94
-----------------
sin r = sin 36.94 /1.33 = 0.4508
r = 26.8 deg
=================
tan (r) = x2 / h = tan 26.8 = 0.505
x2 = 0.505 h ----------- (1)
-------------------------
tan 56.06 = EN / x1
x1 = EN /tan 56.06
EN = 116.35198488189 - h
x1 = [116.35198488189 - h] / 1.486
x1 = 0.673 [116.35198488189 - h] --------- (2)
================
x1+x2 = 11.848481776731 >>> given.
===============
0.505 h +0.673 [116.35198488189 - h] = 11.848481776731
- 0.168 h = - 66.45640404878097
h = real depth of penny = depth of water = 395.5738 cm