An astronomical telescope has focal lengths $100$ & $10$ cm of objective and eyepiece lens respectively when final image is formed at least distance of distinct vision,magnification power of telescope will be,

-15

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-14

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-17

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-19

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Solution

The correct option is A -14
Given focal length of eye piece $f_{e} = 10 c m$
focal length of objective $f_{o} = 100 c m$
Also we know least distance $D = 25 c m$
Magnifying power $M = \frac{- f_{0}}{f_{e}} (1 + \frac{f_{e}}{D})$
$M = - \frac{100}{10} (1 + \frac{10}{25})$
$M = - 14$

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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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respectively. Find its magnifying power, if the final image is formed at a distance of

50 cm

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80 cm

and

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respectively.

A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. 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)?

(b) the final image is formed at the least distance of distinct vision

(25 cm)?

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An astronomical telescope has focal lengths 100 & 10cm of objective and eyepiece lens respectively when final image is formed at least distance of distinct vision,magnification power of telescope will be,

The correct option is A -14Given focal length of eye piecefe=10cmfocal length of objectivefo=100cmAlso we know least distance D=25cm Magnifying power M=−f0fe(1+feD)M=−10010(1+1025)M=−14

An astronomical telescope has focal lengths $100$ & $10$ cm of objective and eyepiece lens respectively when final image is formed at least distance of distinct vision,magnification power of telescope will be,

The correct option is A -14
Given focal length of eye piece $f_{e} = 10 c m$
focal length of objective $f_{o} = 100 c m$
Also we know least distance $D = 25 c m$
Magnifying power $M = \frac{- f_{0}}{f_{e}} (1 + \frac{f_{e}}{D})$
$M = - \frac{100}{10} (1 + \frac{10}{25})$
$M = - 14$