Question

A biologist keeps a specimen of his favorite beetle embedded in a cube of polystyrene plastic. The hapless bug appears to be $2.0\mathrm{cm}$within the plastic. What is the beetle's actual distance beneath the surface?

Short answer

The beetle's actual distance beneath the surface is$s=3.2\mathrm{cm}$

Step-by-step solution

Concepts and Principles

$s^{\mathrm{\prime }}$Scattering from a datum plane can create an image. For refract with such a surface, the subject and imaging lengths are linked by

$s^{\text{'}}=-\frac{n_2}{n_1}s$

The negative sign shows that we are dealing with a virtual image.$n_1$is the index of refraction of the object's medium, $n_2$is the index of refraction of the observer's medium, and distance The object distance , and the image distance being measured from the border.

Given Data

• $n_1=1.59$is the index of refraction of polystyrene plastic.
• $n_2=1.00$is the index of refraction of air
• $s^{\mathrm{\prime }}$=-2.0 cm is the distance of hapless bug image

Required Data

The actual distance $\mathcal{S}$between the beetle and the surface must be determined.

Beetle's actual distance 

Equation connects the beetle's actual distance with its image distance

$s^{\text{'}}=-\left(\frac{n_2}{n_1}\right)s$

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

$s=\left(\frac{1.59}{1.00}\right)\left(2.0\mathrm{cm}\right)$ (Numerically)

$=3.2\mathrm{cm}$