Question

A fish in a flat-sided aquarium sees a can of fish food on the counter. To the fish's eye, the can looks to be $30\mathrm{cm}$ outside the aquarium. What is the actual distance between the can and the aquarium? (You can ignore the thin glass wall of the aquarium.)

Short answer

Thus the actual distance between a can and an aquarium is $22.6\mathrm{cm}$

Step-by-step solution

Using the given data.

Use the relationship between the object's distance, image distance, water's refractive index, and air's refractive index.

$s=\left(\frac{n_{\text{air}}}{n_{\text{water}}}\right)s^{\mathrm{\prime }}$

The $S$is object's distance (actual distance), $s^{\mathrm{\prime }}$the image distance, $n_{air}$the refractive index of air, and $n_{\text{water}}$is the refractive index of water are all given here.

To calculate the actual distance.

Determine the actual distance between the can and the aquarium by using the following formula:

$s=\left(\frac{n_{\text{air}}}{n_{\text{water}}}\right)s^{\mathrm{\prime }}$

Substituting $n_{\text{air}}$, $1.33$for$n_{\text{water}}$, and $30\mathrm{cm}$for $s^{\mathrm{\prime }}$

$s=\left(\frac{1}{1.33}\right)\left(30\mathrm{cm}\right)$

$=22.6\mathrm{cm}$