Determine the position of the image of point P as observed by the...

The apparent shift $\Delta x$ produced by a transparent slab of thickness $t$ and refractive index $\mu$ is given by $\Delta x = t \left(1 - \dfrac{1}{\mu}\right)$.

For multiple slabs, the total shift is the sum of the individual shifts. The air gaps do not contribute to the shift as their refractive index is $1$.

Total shift $\Delta x = \Delta x_1 + \Delta x_2 + \Delta x_3$

$\Delta x = 0.4 \left(1 - \dfrac{1}{1.4}\right) + 0.3 \left(1 - \dfrac{1}{1.3}\right) + 0.2 \left(1 - \dfrac{1}{1.2}\right)$

$\Delta x = 0.4 \left(\dfrac{0.4}{1.4}\right) + 0.3 \left(\dfrac{0.3}{1.3}\right) + 0.2 \left(\dfrac{0.2}{1.2}\right)$

$\Delta x = \dfrac{0.16}{1.4} + \dfrac{0.09}{1.3} + \dfrac{0.04}{1.2}$

$\Delta x = \dfrac{1.6}{14} + \dfrac{0.9}{13} + \dfrac{0.4}{12}$

$\Delta x = \dfrac{0.8}{7} + \dfrac{0.9}{13} + \dfrac{0.1}{3}$

$\Delta x \approx 0.114 + 0.069 + 0.033 = 0.216 \text{ cm}$

Rounding to one decimal place, the total shift is approximately $0.2 \text{ cm}$.

Since the shift is in the direction of the incident light (upwards towards the observer), the image is formed $0.2 \text{ cm}$ above point $P$.

Answer: $0.2 \text{ cm}$ above $P$