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A card sheet divided into squares each of size 1 $\mathrm{mm}^2$ is being viewed at a distance of $9 \mathrm{~cm}$ through a magnifying glass (a converging lens of focal length $9 \mathrm{~cm}$ ) held close to the eye.
(a) What is the magnification produced by the lens? How much is the area of each square in the virtual image?
(b) What is the angular magnification (magnifying power) of the lens?
(c) Is the magnification in (a) equal to the magnifying power in (b)? Explain.
(a) What is the magnification produced by the lens? How much is the area of each square in the virtual image?
(b) What is the angular magnification (magnifying power) of the lens?
(c) Is the magnification in (a) equal to the magnifying power in (b)? Explain.
Solution:
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Verified Answer
(a) Here, u $=-9 \mathrm{~cm}, \mathrm{f}=10 \mathrm{~cm}$
$$
\begin{gathered}
\text { As, } \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \quad \therefore \frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{10}+\frac{1}{-9} \\
=\frac{9-10}{90}=-\frac{1}{90} \quad \therefore v=-90 .
\end{gathered}
$$
$\therefore$ Magnification $=\frac{\mathrm{v}}{\mathrm{u}}=\frac{90}{9}=10$.
In the virtual image, area of each square $=(10 \times 10 \times 1) \mathrm{mm}^2=100 \mathrm{~mm}^2=1 \mathrm{~cm}^2$.
(b) Magnifying power of the lens $=\frac{\mathrm{d}}{\mathrm{u}}=\frac{25}{9}=2.8$.
(c) No, The magnification produced by a lens and its magnifying power are different. The magnitude of magnification is given by $\frac{\mathrm{v}}{\mathrm{u}}$ and magnifying power is $\frac{25}{u}$. only when the image is located at the near point (i.e. $|\mathrm{v}|=25 \mathrm{~cm}$ ) are the two quantities equal.
$$
\begin{gathered}
\text { As, } \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \quad \therefore \frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{10}+\frac{1}{-9} \\
=\frac{9-10}{90}=-\frac{1}{90} \quad \therefore v=-90 .
\end{gathered}
$$
$\therefore$ Magnification $=\frac{\mathrm{v}}{\mathrm{u}}=\frac{90}{9}=10$.
In the virtual image, area of each square $=(10 \times 10 \times 1) \mathrm{mm}^2=100 \mathrm{~mm}^2=1 \mathrm{~cm}^2$.
(b) Magnifying power of the lens $=\frac{\mathrm{d}}{\mathrm{u}}=\frac{25}{9}=2.8$.
(c) No, The magnification produced by a lens and its magnifying power are different. The magnitude of magnification is given by $\frac{\mathrm{v}}{\mathrm{u}}$ and magnifying power is $\frac{25}{u}$. only when the image is located at the near point (i.e. $|\mathrm{v}|=25 \mathrm{~cm}$ ) are the two quantities equal.
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