A small object is located at the centre of the bottom of a...

Let the real depth of the vessel be $H = 4 \text{ cm}$ and its radius be $R = 3 \text{ cm}$.

The object is at the centre of the bottom. A ray travelling from the object to the corner of the vessel has an angle of incidence $i$ at the water surface. From the geometry of the vessel:

$\tan i = \dfrac{R}{H} = \dfrac{3}{4}$

This gives $\sin i = \dfrac{3}{5}$.

Applying Snell's law at the corner of the vessel:

$n \sin i = 1 \cdot \sin r$

$\dfrac{4}{3} \times \dfrac{3}{5} = \sin r$

$\sin r = \dfrac{4}{5}$

This implies $\tan r = \dfrac{4}{3}$.

The refracted ray emerges at an angle $r$ with the normal. When this ray is traced backward, it intersects the normal (the axis of the vessel) at the apparent depth $h'$. From the geometry of the backward-traced ray:

$\tan r = \dfrac{R}{h'}$

$\dfrac{4}{3} = \dfrac{3}{h'}$

$h' = \dfrac{9}{4} = 2.25 \text{ cm}$

The ratio of the real depth to the apparent depth under these assumptions is:

$\text{Ratio} = \dfrac{H}{h'} = \dfrac{4}{2.25} = \dfrac{16}{9} \approx 1.78$