At 10:00hrs a 1.5- m-long vertical stick in air casts a shadow 1.4 m long. If the same stick is placed at 10:00hrs in air in a flat bottomed pool of salt water half the height of the stick, how long is the shadow on the floor of the pool? (For this pool, n = 1.44.)

Question

I know this problem was explained to someone else who obviously has the same set of questions as me. And I tried the steps multiple times and I keep coming up with .404 m, which is incorrect. Can someone walk me through what I'm doing wrong?

Answer 1

The angle of incidence of light hitting the water is arctan 1.4/1.5 = 43.02 degrees

Since the time of day remains the same, the angle of incidence when the shadow is cast on water is the same. The angle of refraction is then given by
1.44 sin R = sin I
sin R = 0.4378
R = 28.28 degrees

The pool depth D is stated to be 0.75 m.

Now consider what happens to a ray of sunlight passing over the top of the stick. It reaches a point 1.4 m beyond the position of the stick at the surface of the pool, and then penetrates the pool, reaching the bottom
at a distance D tan R (0.404 m) beyond 1.4 m. The 0.404 m agrees with what you got, but that is NOT the length of the shadow at the pool bottom.

The total length of the shadow at the bottom of the pool is the sum of 1.4 m and 0.404.

We should keep two significant figures and call it 1.8 m

In my previous answer to this question, I made the same mistake. I forgot to add the length of the shaow at the surface of the water.

Answer 2

My homework site does not accept 1.8 m. What are we doing wrong??

Answer 3

I was calculating the distance of the end of the shadow from a point directly below the stick. The bottom end of the shadow is displaced the same as it is at the surface, 0.404 m. The length of the shadow at the bottom of the pool is the same as it is at the surface: 1.4 m. This is because the rays that pass by the bottom and the top of the pool remain parallel.

I apologize for misleading you.