Someone with a near point P_n of 25 cm views a thimble through a simple magnifying lens of the focal length 10 cm by placing the lens near his eye. What is the angular magnification of the thimble if it is positioned so that its image appears at (a) P_n and (b) infinity?

$\begin{array}{rcl}f& =& 10\text{cm}\\ P_n& =& 25\text{cm}\\ & & \end{array}$

According to the given condition, the image appears at P_n;fromthat, find the object distance and, using the formula of the angular magnification, calculate the magnification when the image appears atP_n. Similarly, calculate the magnification when the image appears at ∞.

The formulas are as follows:

$$\begin{gathered} \frac{\mathbf{1}}{\mathbf{f}}\mathbf{=}\left(\frac{1}{p}+\frac{1}{i}\right) \\ \mathit{m}\mathbf{=}\frac{\mathbf{}{\mathbf{\theta }}^{\mathbf{\text{'}}}}{\mathbf{\theta }} \end{gathered}$$

Where p is the pole, f is the focal length, and i is the image distance.

(a)

Without the magnifier, we can write the angle from figure 34-19,

$\theta =h/P_n$

Where h is the original height of the object.

With magnifier,

$i=-\left|i\right|=-P_n$

So, now, according to the formula,

$\begin{array}{rcl}\frac{1}{f}& =& \left(\frac{1}{P}+\frac{1}{i}\right)\\ \frac{1}{P}& =& \left(\frac{1}{f}+\frac{1}{P_n}\right)\\ P& =& \frac{f{P}_n}{f+P_n}\\ & & \end{array}$

So, now the angular magnification is,

$\begin{array}{rcl}{m}_{\theta }& =& \frac{{\theta }^{\text{'}}}{\theta }\\ m_{\theta }& =& \frac{h/P}{h/P_n}\\ m_{\theta }& =& \frac{P_n}{P}\\ & & \end{array}$

By substituting the value,

$\begin{array}{rcl}{m}_{\theta }& =& \frac{P_n\left(f+P_n\right)}{f{P}_n}\\ m_{\theta }& =& \frac{\left(f+P_n\right)}{f}\\ m_{\theta }& =& 1+\frac{P_n}{f}\\ m_{\theta }& =& 1+\frac{25}{10}\\ & & \end{array}$

$m_{\theta }=3.5$

Therefore, the angular magnification of the image that appears at P_n is 3.5.

(b)

Now, consider the image appears at infinity so that,

$\begin{array}{rcl}i& =& -\left|i\right|=\infty \\ \frac{1}{f}& =& \left(\frac{1}{P}+\frac{1}{i}\right)\\ \frac{1}{P}& =& \frac{1}{f}\\ P& =& f\end{array}$

So, the magnification is,

$\begin{array}{rcl}{m}_{\theta }& =& \frac{h/P}{h/P_n}\\ m_{\theta }& =& \frac{P_n}{P}\\ m_{\theta }& =& \frac{P_n}{f}\\ m_{\theta }& =& \frac{25}{10}\end{array}$

Therefore, the angular magnification of the image that appears at infinity is 2.5.

Using the corresponding formula, find the angular magnification of the image by a simple magnifying lens for different object distances.