The base of a vessel is a plane mirror. Water (refractive index 4/3) of depth 8 cm lies on the mirror and is covered by a floating oil layer…

For near-normal viewing a layer of thickness $t$ and index $\mu$ appears to have thickness $t/\mu$. The mirror therefore appears at depth $\dfrac{8}{4/3}+\dfrac{6}{1.5} = 6+4 = 10\ \mathrm{cm}$ below the oil surface. The mirror forms an image of the object as far behind it as the object is in front, in apparent terms: the object is $5\ \mathrm{cm}$ above the surface, i.e. $5+10 = 15\ \mathrm{cm}$ in front of the apparent mirror. The image therefore appears $15\ \mathrm{cm}$ behind the apparent mirror, at a depth $10+15=25\ \mathrm{cm}$ below the oil surface. Ignoring the refraction gives the trap $5+2(14)=33$ cm.