91Ó°ÊÓ

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 \( - 11\).