What is the Approximate Index of Refraction of a Transparent Liquid?

[Perrin]

Homework Statement

The original problem:

A plane mirror is placed horizontally at the bottom of a transparent liquid of 10 cm depth. When one looks at the image of a small object floating at the surface of the liquid right above the object, the image is seen at 14 cm below the surface of the liquid. Find the approximate value of the index of refraction of the liquid.

Homework Equations

I used Snell's formula:
sin(alpha1)/sin(alpha2) = n2/n1

The Attempt at a Solution

http://www.dotcore.co.il/refraction_problem.jpg
I really hope this image is clear enough, unfortunately that's the top of my Gimp skills...

I said:
sin(alpha1)/sin(alpha2) = n2 (n1 = 1 because it's the index of refraction of air)

sin(alpha1) = BF/AF, sin(alpha2)=DC/AC
DC=FB=x, because the image as seen by the observer as result of the refraction is right above the original image
So:
n2 = (x/AF)/(x/AD) = AD/AF

According to the Pythagorean theorem:
AD = sqrt(x^2 + 20^2)
AF = sqrt(x^2 + 14^2)

Now to my problem:
Certainly, n2 depends on x. I remember my textbook saying that a point below the surface of a material with a higher index of refraction does not create a clear image, is that the reason of the dependency of n2 on x?
The book also said that for small distances, the image can be seen clearly. Does that mean that I can plug in a small value for x to get the correct answer? I tried plugging in x=0, and I got a correct answer. Is it valid to do that? Is that the reason the problem states "Find the approximate value"?

Thanks in advance for the help!
Perrin.

 

[LowlyPion]

Pretty much that looks like it.

For a small angle I think you can reasonably take AF ≈ AB and AC ≈ AD.

 

[Perrin]

OK, thanks for affirming my answer! I can sleep peacefully tonight :D