BRAVO!

That's correct.

You know the apparent depth d_i (the depth of the "image") is given by

 

(this assumes we are looking from the air with n₁ = 1.0, which is usually the case!)

Now we can just "plug in" the numbers,

- 0.75 m = - d_o (¹/_1.33)

The image distance d_i is negative, d_i = - 0.75 m, because this is a "virtual image". The light appears to come from there but it does not actually pass through that point.

- 0.75 m = - d_o (¹/_1.33)

d_o = (0.75 m) (1.33)

d_o = 1.00 m

That is, the "object", your toes, is(/are) 1.0 m below the water's surface.

(c) 2002, Doug Davis; all rights reserved.