A simple magnifier of focal length $\mathit{f}$is placed near the eye of someone whose near point ${\mathit{P}}_{\mathbf{n}}$ is$\mathbf{25}\mathbf{ }\mathbf{cm}$ . An object is positioned so that its image in the magnifier appears at${\mathit{P}}_{\mathbf{n}}$. (a) What is the angular magnification of the magnifier? (b) What is the angular magnification if the object is moved so that its image appears at infinity? For $\mathit{f}\mathbf{=}\mathbf{10}\mathbf{ }\mathbf{cm}$, evaluate the angular magnifications of (c) the situation in (a) and (d) the situation in (b). (Viewing an image at${\mathit{P}}_{\mathbf{n}}$requires effort by muscles in the eye, whereas viewing an image at infinity requires no such effort for many people.)

Near point of the person is, $P_n=25.0 \mathrm{cm}$.

Focal length of the lens is, $f=10 \mathrm{cm}$.

Angular magnification is the ratio of the angular height with magnifier to the angular height without magnifier. For virtual image,$\mathit{i}\mathbf{<}\mathbf{0}$Using this, we can find the required quantities.

Formula:

Angular height$\mathit{\theta }\mathbf{=}\frac{\mathbf{h}}{{\mathbf{P}}_{\mathbf{n}}}$

Angular height with magnifier${\mathit{\theta }}^{\mathbf{\text{'}}}\mathbf{=}\frac{\mathbf{h}}{{\mathbf{P}}_{\mathbf{n}}}$

Angular magnification${\mathit{m}}_{\mathbf{\theta }}\mathbf{=}\frac{\mathbf{\theta }\mathbf{\text{'}}}{\mathbf{\theta }}$

Lens equation $\frac{\mathbf{1}}{\mathbf{P}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{i}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{f}}$

We have,

The angular height without magnifier is, $\theta =\frac{h}{P_n}$

And angular height with magnifier is, ${\theta }^{\text{'}}=\frac{h}{P}$

Now, we can find the position of the virtual image formed by the lens.

Since

$\frac{1}{P}=\frac{1}{f}-\frac{1}{i}$

As the image is virtual, $i=-\left|i\right|=P_n$

Therefore,

$\frac{1}{P}=\frac{1}{f}+\frac{1}{\left|i\right|}$

$=\frac{1}{f}+\frac{1}{P_n}$

Then the angular magnification is

$m_{\theta }=\frac{\theta \text{'}}{\theta }$

$\begin{array}{c}=\left(\frac{h/P}{h/P_n}\right)\\ =\left(\frac{1/P}{1/P_n}\right)\end{array}$

$\begin{array}{c}{m}_{\theta }=\frac{\left(\frac{1}{f}+\frac{1}{P_n}\right)}{\frac{1}{P_n}}\\ =\frac{P_n}{f}+1\\ =\frac{25 \mathrm{cm}}{f}+1\end{array}$

For focal length $f$, the angular magnification of the magnifier is,$m_{\theta }=1+\frac{25 \mathrm{cm}}{f}$

If object is moved so that its image appears at infinity,

then $i=-\left|i\right|\to -\infty$.

$\begin{array}{c}\frac{1}{P}=\frac{1}{f}+\frac{1}{\left|i\right|}\\ =\frac{1}{f}+\frac{1}{\infty }\\ =\frac{1}{f}\end{array}$

Then the angular magnification is

$m_{\theta }^{\text{'}}=\frac{\theta \text{'}}{\theta }$

$=\left(\frac{h/p}{h/p_n}\right)$

$=\left(\frac{1/p}{1/p_n}\right)$

$\begin{array}{c}{m}_{\theta }^{\text{'}}=\frac{P_n}{f}\\ =\frac{25 \mathrm{cm}}{f}\end{array}$

For focal length$f,$ the angular magnification if the object is moved so that its image appears at infinity is,

$m_{\theta }^{\text{'}}=\frac{25 \mathrm{cm}}{f}$

For$f=10 \mathrm{cm}$, the angular magnification of magnifier is

$\begin{array}{c}{m}_{\theta }=\frac{P_n}{f}+1\\ =\frac{25 \mathrm{cm}}{10 \mathrm{cm}}+1\\ =3.5\end{array}$

For $f=10 \mathrm{cm}$, the angular magnification of the magnifier is $\begin{array}{c}{m}_{\theta }=3.5\end{array}$

For $f=10 \mathrm{cm}$, the angular magnification if the object is moved so that its image appears at infinity is

$\begin{array}{c}{m}_{\theta }^{\text{'}}=\frac{P_n}{f}\\ =\frac{25 \mathrm{cm}}{10 \mathrm{cm}}\\ =2.5\end{array}$

For $f=10 \mathrm{cm}$, the angular magnification if the object is moved so that its image appears at infinity is $m_{\theta }^{\text{'}}=2.5$