Question

A biologist keeps a specimen of his favorite beetle embedded in a cube of polystyrene plastic. The hapless bug appears to be 2.0cm within the plastic. Part A What is the beetle's actual distance beneath the surface? Express your answer using two significant figures..

Answer 1

Concepts and reason

The main concept required to solve this problem is the apparent depth, actual depth and refractive index.

First, write the equation for the relation of the apparent depth, and actual depth. Use this formula and calculate the actual depth of the beetle beneath the surface.

Fundamentals

The equation for the relation for the apparent depth and actual depth is,

$n = \frac{{{d_{{\rm{actual}}}}}}{{{d_{{\rm{apparnt}}}}}}$

Here, ${d_{{\rm{actual}}}}$ is the actual depth, ${d_{{\rm{apparnt}}}}$ is the apparent depth and n is the refractive index of the medium.

The equation for the relation of the actual depth and apparent depth of the beetle is,

$n = \frac{{{d_{{\rm{actual}}}}}}{{{d_{{\rm{apparnt}}}}}}$

Here, ${d_{{\rm{actual}}}}$ is the actual distance of the beetle, ${d_{{\rm{apparnt}}}}$ is the apparent distance of the beetle and n is the refractive index of the cube of polystyrene plastic.

Rearrange the above equation for the actual distance of the beetle, ${d_{{\rm{actual}}}}$ .

${d_{{\rm{actual}}}} = n \times \left( {{d_{{\rm{apparnt}}}}} \right)$

The equation for the actual distance of the beetle that derived in above step is,

${d_{{\rm{actual}}}} = n \times \left( {{d_{{\rm{apparnt}}}}} \right)$

Substitute 1.62 for n, and $2.0{\rm{ cm}}$ for ${d_{{\rm{apparnt}}}}$ in above equation,

$\begin{array}{c}\\{d_{{\rm{actual}}}} = \left( {1.62} \right) \times \left( {2.0{\rm{ cm}}} \right)\\\\ = 3.24{\rm{ cm}}\\\end{array}$

Ans:

The actual distance of the beetle from the surface of the cube is 3.24 cm.