Question

Find the apparent depth of an object O placed at the bottom of a beaker as shown in which two layers of transparent liquids are filled.

$h_1$ = 25 cm $\mu_1$ = 1.5

$h_2$ = 15 cm $\mu_2$ = 2.5

Answer 1

$\frac{68}{3}$ cm

Answer 2

The apparent depth of an object viewed from above through a medium of refractive index $\mu$ at a real depth $h$ is given by $h_{app} = h/\mu$. When the object is viewed through multiple layers of different media, the total apparent depth is the sum of the apparent depths of each layer.

Given: Layer 1: thickness $h_1 = 25$ cm, refractive index $\mu_1 = 1.5$. Layer 2: thickness $h_2 = 15$ cm, refractive index $\mu_2 = 2.5$.

The apparent depth due to the first layer is $h_{1,app} = \frac{h_1}{\mu_1}$. The apparent depth due to the second layer is $h_{2,app} = \frac{h_2}{\mu_2}$.

The total apparent depth is the sum of these apparent depths: $h_{total,app} = h_{1,app} + h_{2,app}$

Calculation: $h_{1,app} = \frac{25 \text{ cm}}{1.5} = \frac{25}{3/2} \text{ cm} = \frac{50}{3} \text{ cm}$ $h_{2,app} = \frac{15 \text{ cm}}{2.5} = \frac{15}{5/2} \text{ cm} = \frac{30}{5} \text{ cm} = 6 \text{ cm}$

$h_{total,app} = \frac{50}{3} \text{ cm} + 6 \text{ cm} = \frac{50}{3} + \frac{18}{3} \text{ cm} = \frac{68}{3} \text{ cm}$

Approximately, $\frac{68}{3} \text{ cm} \approx 22.67 \text{ cm}$.