Find the magnification of the heater element in Example \(25.9\). Note that its large magnitude helps spread out the reflected energy.

Ratio of image distance to the object distance is called magnification.

Given values of image and object distances:

\({d_i} = 3\;\;m\)

\(\begin{aligned}{d_0} &= 27.3\;cm\\ &= 27.3 \cdot \left( {1\;cm \times \dfrac{{1 \times {{10}^{ - 2}}\;m}}{{1\;cm}}} \right)\\ &= 27.3 \times {10^{ - 2}}\;m\end{aligned}\)

\({d_o}\) denotes the distance between the object and the lens, and \({d_i}\) denotes the distance from the lens to the projected image that is in focus.

The magnification of the heater element, we will find from the equation:

\(m = - \dfrac{{{d_i}}}{{{d_o}}}\)

where \(m\) denotes the magnification, \({d_o}\) denotes the distance between the object and the lens, \({d_i}\) denotes the distance from the lens to the projected image that is in focus.

\(\begin{aligned}m &= - \dfrac{{{d_i}}}{{{d_o}}}\\m &= - \dfrac{{300\;cm}}{{27.3\;cm}}\\m &= - 11\end{aligned}\)

Hence, magnification of heater element is \( - 11\).