Question

an astronomical telescope has a length of 105cm,and its magnification is 6 determine the power of objective and eye piece?

Answer 1

Answer on Question #41799, Physics, Optics

An astronomical telescope has a length of $105\mathrm{cm}$ , and its magnification is 6 determine the power of objective and eye piece?

Solution:

Given:

$L = 105\mathrm{cm}$

$M = 6$

$f_{o} = ?$

$f_{e} = ?$

Focal length of objective $= f_{o}$

Focal length of eyepiece $= f_{e}$

When the final image is at infinity, then magnification is

$$M = \frac {f _ {o}}{f _ {e}} = 6$$

and length of the telescope is

$$L = f _ {o} + f _ {e} = 1 0 5 \mathrm {c m}$$

Thus, from first equation

$$f _ {o} = 6 f _ {e}$$

From second equation

$$6 f _ {e} + f _ {e} = 1 0 5$$ $$7 f _ {e} = 1 0 5$$ $$f _ {e} = \frac {1 0 5}{7} = 1 5 \mathrm {c m}$$ $$f _ {o} = 6 \cdot 1 = 9 0 \mathrm {c m}$$

The power of a lens is defined as the reciprocal of its focal length in meters.

Thus, the powers of objective and eye piece are

$$P _ {o} = \frac {1}{f _ {o}} = \frac {1}{0 . 9} = 1. 1 1 \text {d i o p t e r s}$$ $$P _ {e} = \frac {1}{f _ {e}} = \frac {1}{0 . 1 5} = 6. 6 7 \text {d i o p t e r s}$$

Answer. $P_{o} = 1.11$ diopters, $P_{e} = 6.67$ diopters.

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