Concept used

When a layered liquid system is viewed from above, the apparent depth is the sum of the apparent thicknesses of each layer: $d_{\text{apparent}} = \sum \frac{t_i}{\mu_i}$ where $t_i$ is real thickness and $\mu_i$ is refractive index.

The real depth of the bottom is: $d_{\text{real}} = \sum t_i$

Hence, the apparent shift is: $\text{shift} = d_{\text{real}} - d_{\text{apparent}}$

Given data

• First liquid: refractive index $\mu_1 = 1.2$, height $t_1 = 60\,\text{cm}$
• Second liquid: refractive index $\mu_2 = 1.6$, height $t_2 = H\,\text{cm}$
• Apparent shift of bottom = $40\,\text{cm}$

Write real depth

$d_{\text{real}} = 60 + H$

Write apparent depth

$d_{\text{apparent}} = \frac{60}{1.2} + \frac{H}{1.6}$

Now, $\frac{60}{1.2} = 50, \qquad \frac{H}{1.6} = \frac{5H}{8}$

So, $d_{\text{apparent}} = 50 + \frac{5H}{8}$

Use apparent shift = 40 cm

$d_{\text{real}} - d_{\text{apparent}} = 40$

$\left(60 + H\right) - \left(50 + \frac{5H}{8}\right) = 40$

Simplify: $10 + H - \frac{5H}{8} = 40$

$10 + \frac{3H}{8} = 40$

$\frac{3H}{8} = 30$

$3H = 240$

$H = 80\,\text{cm}$

Final answer

$\boxed{80}$