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$\begin{array}{l}{\text { You want to view an insect } 2.00 \mathrm{mm} \text { in length through a }} \\ {\text { magnifier. If the insect is to be at the focal point of the magnifier, }} \\ {\text { what focal length will give the image of the insect an angular size }} \\ {\text { of } 0.025 \text { radian? }}\end{array}$

$y=2 m m$

$\theta^{\prime}=0.025 \text { radian }$

Thin Lens $\Rightarrow \frac{-s'}{s} = \frac{y'}{y} \Rightarrow$$\left|\frac{y^{\prime}}{s^{\prime}}\right|=\left|\frac{y}{s}\right|$

$\theta={\theta'}\Rightarrow \theta=\frac{y}{f} \Rightarrow ∴ f=\frac{y}{\theta}=\frac{2}{0.025}$

$∴ f=80 m m=8 cm$

$\begin{array}{l}{\text { Resolution of a Microscope. The image formed by a }} \\ {\text { microscope objective with a focal length of } 5.00 \mathrm{mm} \text { is } 160 \mathrm{mm} \text { from }} \\ {\text { its second focal point. The eyepiece has a focal length of } 26.0 \mathrm{mm} \text { . }} \\ {\text { (a) What is the angular magnification of the microscope? (b) The }} \\ {\text { unaided eye can distinguish two points at its near point as separate }} \\ {\text { if they are about } 0.10 \mathrm{mm} \text { apart. What is the minimum separation }} \\ {\text { between two points that can be observed (or resolved) through this }} \\ {\text { microscope? }}\end{array}$

a) $f_{1}=5 \mathrm{mm}$

$S_{1}^{\prime}=160+5=165 \mathrm{mm}$

$f_{2}=26 \mathrm{mm}$

$M=\frac{(25 \mathrm{cm}) \mathrm{S}_{1}^{\prime}}{f_{1} \mathrm{f}_{2}}$$=\frac{(250) * 165}{5 * 26}=317$

b) $\text { distance }=0.1 mm / \text { the minimum separation }$$=\frac{0.1m m}{M}=\frac{0.1m m}{317}$

$=3.15 \times 10^{-9} mm$

$\begin{array}{l}{\text { The eyepiece of a refracting telescope (see Fig. } 34.53 )} \\ {\text { has a focal length of } 9.00 \mathrm{cm} . \text { The distance between objective and }} \\ {\text { eyepiece is } 1.80 \mathrm{m}, \text { and the final image is at infinity. What is the }} \\ {\text { angular magnification of the telescope? }}\end{array}$

$f_{2}=9 \mathrm{cm}$

$f_{1}+f_{2}=1.8 \mathrm{m}$

$∴ f_{1}=1.8 -\left(9 * 10^{-2}\right)=1.71 m$

$M=\frac{-f_{1}}{f_{0}}=\frac{-1.71}{0.09}=-19$

$\text { Telescope }$$\Rightarrow f_{1}>>f_2$