A man with normal near point (25 cm) reads a book with small print using a magnifying glass: a thin convex lens of focal length 5 cm.
- 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?
- What is the maximum and the minimum angular magnification (magnifying power) possible using the above simple microscope?
Exercise
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- For the closest distance,
Focal length of thin convex lens, f = 5 cm
Using the formula,
$\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\therefore \ \frac{1}{\text{u}}=\frac{1}{\text{v}}-\frac{1}{\text{f}}$
$\frac{1}{-25}-\frac{1}{\text{u}}=\frac{1}{5}$
$-\frac{1}{\text{u}}=\frac{1}{5}+\frac{1}{25}$
$-\frac{1}{\text{u}}=\frac{5+1}{25}$
$-\frac{1}{\text{u}}=\frac{6}{25}$
$-\text{u}=\frac{25}{6}$
$\therefore\ \text{u}=-\frac{25}{6}=-4.2 \ \text{cm}$
l.e.. 4.2 cm is the closest distance at which the man can read the book.
For the farthest image,
Image distance, v = $\infty$
Focal length, f = 5 cm
Using the formula,
$\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\frac{1}{\text{u}}=\frac{1}{\text{v}}-\frac{1}{\text{f}}$
$=\frac{1}{\infty}-\frac{1}{5}$
$=-\frac{1}{5}$
$\text{u} = -5 \text{cm}$
which is the object distance
This is the farthest distance at which the man can read the book.
- Normal near point of human eye = 25 cm
Maximum angular magnification is
$\frac{\text{D}}{\text{U}_\text{min}}=\frac{25}{\frac{25}{6}}=6$
Minimum angular magnification is
$\frac{\text{D}}{\text{U}_\text{max}}=\frac{25}{5}=5.$
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