Name: Angular Magnification and Magnifying Power
Uploaded: 2019-07-19
Duration: 7 min 43 s
Description: Angular magnification can be used to assess the size of the object as it appears to the eye. To learn more about angular magnification produced by an optical...

Question

The objective of a telescope focussed for the least distance of distinct vision forms the image $0.75 m$ behind it. The distance between the two lenses is $0.80 m$ . What is the magnifying power of the telescope? For seeing with relaxed eye, by how much distance will the eye-piece have to be moved back? What will be the magnifying power then?

Accepted Answer

Given Distance between the lenses = $L = 80 cm$

Distance of image from the objective (focus) $f_{o_{}} = 75 cm$

We know that the distance between the lenses is sum of focal lengths of objective and eyepiece,

$L = f_{o} + f_{e} or 80 = 75 + f_{e} or f_{e} = 5 cm .$ , where $f_{e}$ is the focal length of the eye piece

Magnifying power $m = - \frac{f_{o}}{f_{e}} = \frac{75}{5} = - 15$ .......................(1)

For relaxed eyes, the final image is at the least distance of clear vision i.e $v_{e} = 25 cm$ ,

Applying lens formula for the eyepiece

$\frac{i}{f_{e}} = \frac{1}{v_{e}} - \frac{1}{u} or \frac{1}{5} = \frac{1}{- 25} - \frac{1}{(- u)} or \frac{1}{u} = \frac{- 5 + 1}{25}$ or $u = - \frac{25}{6} = - 6.25 cm$

So we have to move back the eyepiece= $6.25 ~ 5 = 1.25 cm$ ...............(2)

The magnifying power in this case will be

$m = - \frac{f_{o}}{f_{e}} (1 + \frac{f_{e}}{d}) = - 15 (1 + \frac{(- 5)}{25}) = - 12$ ..........................(3)

Hence, the magnifying power is $- 15$ . The eye lens has to be shifted by $1.25 cm$ and new magnifying power is $- 12$ .

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