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A man with normal near point $(25 \mathrm{~cm})$ reads a book with small print using a magnifying glass: a thin convex lens of focal length $5 \mathrm{~cm}$.
(a) What is the closest and the farthest distance at which he should keep the lens from the page so that he can read the book when viewing through the magnifying glass?
(b) What is the maximum and the minimum angular magnification (magnification power) possible using the above simple microscope?
(a) What is the closest and the farthest distance at which he should keep the lens from the page so that he can read the book when viewing through the magnifying glass?
(b) What is the maximum and the minimum angular magnification (magnification power) possible using the above simple microscope?
Solution:
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Verified Answer
(a) At closest distance of the object the image is formed at least distance of distinct vision and eye is most strained.
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ or $\frac{1}{-25}-\frac{1}{u}=\frac{1}{5}$
$$
-\frac{1}{u}=\frac{1}{5}+\frac{1}{25}=\frac{5+1}{25}
$$
or $u=-\frac{25}{6} \mathrm{~cm}=-4.2 \mathrm{~cm}$
At farthest distance of the object the image is formed at $\infty$ and eye is most relaxed.
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ or $\frac{1}{-\infty}-\frac{1}{u}=\frac{1}{5}$
(b) Maximum angular magnification
$$
m_{\max }=\left[1+\frac{D}{f_e}\right] \text { or } \frac{D}{u_{\min }}=\left[1+\frac{25}{5}\right]
$$
or $\frac{25}{\frac{25}{6}}$
Minimum angular magnification
$$
m_{\min }=\frac{D}{f_e} \text { or } \frac{D}{u_{\min }}=\frac{25}{5}=5
$$
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ or $\frac{1}{-25}-\frac{1}{u}=\frac{1}{5}$
$$
-\frac{1}{u}=\frac{1}{5}+\frac{1}{25}=\frac{5+1}{25}
$$
or $u=-\frac{25}{6} \mathrm{~cm}=-4.2 \mathrm{~cm}$
At farthest distance of the object the image is formed at $\infty$ and eye is most relaxed.
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ or $\frac{1}{-\infty}-\frac{1}{u}=\frac{1}{5}$
(b) Maximum angular magnification
$$
m_{\max }=\left[1+\frac{D}{f_e}\right] \text { or } \frac{D}{u_{\min }}=\left[1+\frac{25}{5}\right]
$$
or $\frac{25}{\frac{25}{6}}$
Minimum angular magnification
$$
m_{\min }=\frac{D}{f_e} \text { or } \frac{D}{u_{\min }}=\frac{25}{5}=5
$$
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