Question

You are playing a game where you drop a coin into a water tank and try to land it on a target. You often find this game at fast food places as part of a fundraiser. The sides of the tank are flat and the target is 6in from the side of the tank. Your eye is 9in from the side of the tank. Water has an index of refraction of 4/3. If you drop the coin accurately and it falls straight down to the location where the target appears to be, how far are you off?

Answer 1

Answer:

1.5 inch in the front of the target

Step-by-step explanation:

distance of the object, y = 6 in.

Since the surface is flat, then the radius of curvature has to be infinity.

The incident index is n(i) = 4/3 and the transmitted index is n(t) = 1

The single interface equation is

[n(i) / y + n(t) / y^i] = [n(t) -n(i)] / r

When we substitute the appropriate values, we have

1.333/6 + 1/y^i = 0

1/y^i = -1.3333/6

y^i = -6/1.333

y^i = -4.5

thus, the image distance is -4.5 in

Therefore, the coin falls 1.5 inch in the front of the target