Question

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

Short answer

Magnification of heater element is \( - 11\).

Step-by-step solution

Definition of magnification

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

Information Provided

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.

Calculating magnification of heater element

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 \( - 11\).