Question

A tank containing a slab of glass 10cm thick of n=1.6 above this is a depth of 5cm of liquid of n=1.5 and upon this floats 5cm of water of n=4÷3. To an observer looking down form above what is the apparent position of the object at the bottom of the tank

Answer 1

The apparent position of the object at the bottom of the tank:

$$D=\left(1-\frac{1}{\mu_1}\right)t_1+\left(1-\frac{1}{\mu_2}\right)t_2+\left(1-\frac{1}{\mu_3}\right)t_3\\D=\left(1-\frac{1}{1.6}\right)10+\left(1-\frac{3}{4}\right)5+\left(1-\frac{1}{1.5}\right)5\\D=6.67\ cm$$