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Standard XII

Physics

Angular Magnification in Telescope

Focal length ...

Question

Focal length of the objective of a terrestrial telescope is $80 cm$ and it is adjusted for parallel rays, then its magnifying power is $20$ . If the focal length of erecting lens is $20 cm,$ then full length of telescope will be ( in $cm$ )

164

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124

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100

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84

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Solution

The correct option is A $164$
Given that
$m = 20; f_{o} = 80 cm; f = 20 cm (erecting lens)$
Magnification
$m = \frac{f_{o}}{f_{e}} \Rightarrow 20 = \frac{80}{f_{e}} \Rightarrow f_{e} = \frac{80}{20} = 4 cm (eye piece)$
Length of terrestial telescope
$L = f o + 4 f + f e$
$L = 80 + (4 \times 20) + 4$
$L = 80 + 80 + 4$
$L = 164 cm$

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Focal length of the objective of a terrestrial telescope is $80 cm$ and it is adjusted for parallel rays, then its magnifying power is $20$ . If the focal length of erecting lens is $20 cm,$ then full length of telescope will be ( in $cm$ )

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Focal length of the objective of a terrestrial telescope is 80 cm and it is adjusted for parallel rays, then its magnifying power is 20. If the focal length of erecting lens is 20 cm, then full length of telescope will be ( in cm)

The correct option is A 164Given that m=20 ;fo=80 cm ;f=20 cm (erecting lens) Magnification m=fofe⇒20=80fe⇒fe=8020=4 cm (eye piece) Length of terrestial telescope L=fo+4f+fe L=80+(4×20)+4 L=80+80+4 L=164 cm

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Focal length of the objective of a terrestrial telescope is $80 cm$ and it is adjusted for parallel rays, then its magnifying power is $20$ . If the focal length of erecting lens is $20 cm,$ then full length of telescope will be ( in $cm$ )