Chapter 26 - Geometrical Optics - Problems and Conceptual Exercises - Page 942: 61

$4.7ft$

We know that $\frac{\theta_2}{\theta_1}=\frac{n_{water}}{n_{air}}$ $\frac{\theta_2}{\theta_1}=\frac{1.33}{1.00}=1.33$ Now we can find the apparent depth as follows: $d_{app}=\frac{d}{angular \space size}$ We plug in the known values to obtain: $d_{app}=\frac{6.5ft}{1.33}$ $d_{app}=4.7ft$