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$$
\begin{array}{l}{\text { A converging lens forms an image of an } 8.00 \text { -mm-tall }} \\ {\text { real object. The image is } 12.0 \mathrm{cm} \text { to the left of the lens, } 3.40 \mathrm{cm}} \\ {\text { tall, and erect. What is the focal length of the lens? Where is the }} \\ {\text { object located? }}\end{array}
$$

$$
y=8 m m
$$

$$
s^{\prime}=12 \mathrm{cm}
$$

$$
y^{\prime}=3.4 \mathrm{cm}
$$

$$
s=? ? > f=? ?
$$

$$
m=\frac{-s'}{s}=\frac{y'}{y}
$$

\(∴ \frac{-(-12)}{s} =\frac{3 \cdot 4}{8 * 10^{-1}} \Rightarrow S=2.82 cm \)

\( \frac{1}{s}+\frac{1}{s^{\prime}}=\frac{1}{f} \Rightarrow ∴ \frac{1}{2.82}+\frac{1}{(-12)}=\frac{1}{f} \)

\(∴ f=2.69 cm \)

$$
\begin{array}{l}{\text { A double-convex thin lens has surfaces with equal radii }} \\ {\text { of curvature of magnitude } 2.50 \mathrm{cm} . \text { Looking through this lens, }} \\ {\text { you observe that it forms an image of a very distant tree at a dis- }} \\ {\text { tance of } 1.87 \mathrm{cm} \text { from the lens. What is the index of refraction of }} \\ {\text { the lens? }}\end{array}
$$

$$
R=2.5 \mathrm{cm}
$$

$$
S=\infty
$$

$$
S^{\prime}=1.87 \mathrm{cm}
$$

$$
n=? ?
$$

$$
f=1.87 cm
$$

\( R_{1}=+R\quad ,\quad ∴ R_{2}=-R\quad , \quad R=2.5 \mathrm{cm} \)

$$
\frac{1}{f}=(n-1)\left(\frac{1}{R}-\frac{1}{(-R)}\right)
$$

\(∴ =\frac{1}{f}=\frac{2(n-1)}{R} \Rightarrow ∴ 2 f= \frac{R}{(n-1)} \Rightarrow n=\frac{R}{2 f}+1 \)

\(∴ n =\frac{2.5}{2*1.87}+1=1.67 \)

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