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

The actual distance between the can and the aquarium

$s=22.6\mathrm{cm}$

Step-by-step solution

Concepts and Principles

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.

Given Data

• 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:$n_2=1.33$

Required Data

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

Solution

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)