Magnifying power (in magnitude) of telescope for viewing distant objects when the telescope is in normal adjustment (final image is at infinity) :

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Solution

The correct option is D 28
the magnification of a telescope is given by,
m = $\frac{f_{o}}{f_{e}}$

Hence, $m = \frac{140}{5}$

$m = 28$

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Similar questions

Q. A small telescope has an objective lens of focal length 140cm andan eyepiece of focal length 5.0cm. What is the magnifying power ofthe telescope for viewing distant objects when (a) the telescope is in normal adjustment (i.e., when the final imageis at infinity)? (b) the final image is formed at the least distance of distinct vision(25cm)?

Answer the following questions:

(a) For the telescope described in Exercise (a)[A small telescope has an objective lens of focal length

140 c m

and an eyepiece of focal length

5.0 c m

. What is the magnifying power of the telescope for viewing distant objects when

(a) the telescope is in normal adjustment (i.e., when the final image is at infinity)?],

What is the separation between the objective lens and the eyepiece?

(b) If this telescope is used to view a

100 m

tall tower

3 k m

away, what is the height of the image of the tower formed by the objective lens?

25 c m

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A small telescope has an objective lens of focal length 150 cm and an eye-piece of focal length 5 cm. What is the magnifying power of the telescope for viewing distant objects in normal adjustment?
If this telescope is used to view a 100 m tall tower 3 km away, what is the height of the image of the tower formed by the objective lens?

Q. A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. The magnifying power of the telescope for viewing distant objects when the final image is formed at the least distance of distinct vision 25 cm will be

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Magnifying power (in magnitude) of telescope for viewing distant objects when the telescope is in normal adjustment (final image is at infinity) :

The correct option is D 28the magnification of a telescope is given by,m = fofeHence, m=1405m=28

The correct option is D 28
the magnification of a telescope is given by,
m = $\frac{f_{o}}{f_{e}}$

Hence, $m = \frac{140}{5}$

$m = 28$