How Does Refraction Affect Apparent Depth in a Layered Liquid System?

[larkinfan11]

Homework Statement

A beaker has a height of 40.0 cm. The lower half of the beaker is filled with water (n = 1.333), and the upper half is filled with oil (n = 1.48). To a person looking down into the beaker from above, what is the apparent depth of the bottom?

Homework Equations

Apparent Depth= d(n(observer)/n(object))

The Attempt at a Solution

I believe this is a two part problem, so I broke it down into that. First I solved it from the perspect of looking observing from the oil level to the water level and calculated this:

Apparent Depth= 20(1.48/1.333)=22.2

I used 20 as the distance between the water and the bottom because it was half of the size of the beaker. That may be incorrect, however.

I then used the apparent depth that I calcuated above to solve the second part from the aspect of looking at this from the air to the oil and calculated this:

22.2= d(1/1.48)
d=32.856

I then added the two together to get 55.056 cm as my total apparent depth, but the answer is incorrect. What am I doing wrong?

 

[Andrew Mason]

The apparent depth should be shallower than the actual.

If the oil was not there, the depth would appear to be 20 + 20/1.33 = 35 cm. With the oil, what would it be?

AM

 

[larkinfan11]

The apparent depth should be shallower than the actual.

If the oil was not there, the depth would appear to be 20 + 20/1.33 = 35 cm. With the oil, what would it be?

AM

I think I see what you're saying. With the oil there, the equation should look like this: 20/1.48 + 20(1.48/1.333) Am I on the right track?

 

[Andrew Mason]

I think I see what you're saying. With the oil there, the equation should look like this: 20/1.48 + 20(1.48/1.333) Am I on the right track?

Not quite. Just consider the depth of an object at 20 cm below the oil only. What would the apparent depth be? What if the oil extended to 40 cm? (hint: 40/1.48 = 20/1.48 + 20/1.48)? Now think of those last 20 cm in water.

AM