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.)

Waves out of a flat plane are going to be accustomed buildan image . During refract with such a surface, the source and photo lengths is correlated.

$s^{\mathrm{\prime }}=-\frac{n_2}{n_1}s$

The sign indicates that we are handling a reflexion. $n_1$ is that the index of refraction of the object's medium, $n_2$ is that the index of refraction of the observer's medium, and distance the item distance , and so the image distance being measured from the border.

• The can's image distance is: $s^{\mathrm{\prime }}=30\mathrm{cm}$
• The index of refraction of air is: $n_1=1.00$.
• The index of refraction of air is:$n_2=1.33$

We're speculated to determine how far the can is from the aquarium .

Equation connects the distances between the can and its picture

$s^{\mathrm{\prime }}=-\frac{n_2}{n_1}s$

$s=-\left(\frac{n_1}{n_2}\right)s^{\mathrm{\prime }}$

$s=-\left(\frac{1.00}{1.33}\right)(-30\mathrm{cm})$ (Numerically)