How is the working of a telescope different from that of a microscope?
The focal lengths of the objective and eyepiece of a microscope are 1.25 cm and 5 cm respectively. Find the position of the object relative to the objective in order to obtain an angular magnification of 30 in normal adjustment.
The focal lengths of the objective and eyepiece of a microscope are 1.25 cm and 5 cm respectively. Find the position of the object relative to the objective in order to obtain an angular magnification of 30 in normal adjustment.
CBSE DELHI - SET 1 2012
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Working differences:
Given: $\text{f}_{o} = 1.25\text{cm}$
$\text{f}_{e} = 5 \text{cm}$
Angular magnification m= 30
Now, $m = m_e \times m_o$
In normal adjustment, angular magnification of eyepiece
$\text{m}_{e} = \text{d} / \text{f}_{e} = + \frac{25}{5} = 5 $
Hence, $m_o = 6$
But $\text{m}_{o} = \frac{\text{v}_{o}}{\text{u}_{o}} = > -6 =\frac{\text{v}_{o}}{\text{u}_{o}}$
$ = > \text{v}_{o} = - 6 \text{u}_{o}$
Applying lens equation to the objective lens:
$\frac{1}{\text{f}_{o}} = \frac{1}{\text{v}_{o}} - \frac{1}{\text{u}_{o}}$
$\frac{1}{1.25} = \frac{1}{-6\text{u}_{o}} - \frac{1}{\text{u}_{o}}$
$\frac{1}{1.25} = \frac{-1-6}{6\text{u}_{o}}$
$6\text{u}_{o} = 1.25\times(-7)$
$\text{u}_{o} = \frac{-1.25\times7}{6}\text{ cm}$
$ = - 1.46\text{cm}.$
- Objective of a telescope forms the image of a very far off object at, or within, the focus of its eyepiece. The microscope does the same for a small object kept just beyond the focus of its objective.
- The final image formed by a telescope is magnified relative to its size as seen by the unaided eye while the final image formed by a microscope is magnified relative to its absolute size.
- The objective of a telescope has large focal length & large aperture while the corresponding for a microscope have very small values.
| Telescope | Microscope |
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$\text{f}_{e} = 5 \text{cm}$
Angular magnification m= 30
Now, $m = m_e \times m_o$
In normal adjustment, angular magnification of eyepiece
$\text{m}_{e} = \text{d} / \text{f}_{e} = + \frac{25}{5} = 5 $
Hence, $m_o = 6$
But $\text{m}_{o} = \frac{\text{v}_{o}}{\text{u}_{o}} = > -6 =\frac{\text{v}_{o}}{\text{u}_{o}}$
$ = > \text{v}_{o} = - 6 \text{u}_{o}$
Applying lens equation to the objective lens:
$\frac{1}{\text{f}_{o}} = \frac{1}{\text{v}_{o}} - \frac{1}{\text{u}_{o}}$
$\frac{1}{1.25} = \frac{1}{-6\text{u}_{o}} - \frac{1}{\text{u}_{o}}$
$\frac{1}{1.25} = \frac{-1-6}{6\text{u}_{o}}$
$6\text{u}_{o} = 1.25\times(-7)$
$\text{u}_{o} = \frac{-1.25\times7}{6}\text{ cm}$
$ = - 1.46\text{cm}.$
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