At 08:00hrs a 2.7- m-long vertical stick in air casts a shadow 1.8 m long. If the same stick is placed at 08:00hrs 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 = 1.64.)

Question

I can't get this for the life of me, I keep getting x= .4852 m
please tell me what i am doing wrong.

Answer 1

You are probably doing the refraction part right, but consider how the stick shadow is formed. It is already 1.8 m long at the surface of the water. You have to add that to the additional lateral distance that the light hitting the top of the pole refracts under water.

See how I answered the other problem that I just posted, which was of the same type.

That is a pretty high sun angle and short shadow for 8 AM, but we have to go with what they give us.

Answer 2

okay so i add the .4852 to the 1.8 and i get 2.2852 but it keeps saying that im wrong.
i don't understand

Answer 3

Isn't light bent toward the normal in the higher refractive medium?

Answer 4

I agree with the .4852 m lateral deflection of the beam at the top of the shadow after it enters the pool. Try answering with two or three significant figures: 2.29 m or 2.3 m. You only know the stick length to two figures, so predictions cannot be more accurate than that.

Answer 5

It's still wrong. I have no idea what to do at this point, but thanks for your help anyway.

Answer 6

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, 0.4852 m. The length of the shadow at the bottom of the pool is the same as it is at the surface: 1.8 m. This is because the rays that pass by the bottom and the top of the pool remain parallel.

I apologize for misleading you.