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The focal lengths of objective lens and eye lens of a Gallelian Telescope are respectively 30 cm and 3.0 cm. Telescope produces virtual, erect image of an object situated far away from it at least distance of distinct vision from the eye lens. In this condition, the Magnifying Power of the Gallelian Telescope should be :

  • Option 1)

    +11.2

  • Option 2)

    -11.2

  • Option 3)

    - 8.8

  • Option 4)

    + 8.8

 

Answers (1)

best_answer

As we have learned

Astronomical Telescope -

m= \frac{-f_{o}}{f_{e}}\left ( 1+\frac{f_{e}}{D} \right )
 

- wherein

f_{o} = focal length of objective

f_{e}= focal length of eyepiece

 

 Given focal length of 

Objective f_0 = 30 cm

eyepiece f_e = 3 cm

M = \frac{f_0}{f_e}(1- \frac{f_e}{D})

= \frac{30}{3}(1- \frac{3}{25})= 10 \times \frac{22}{25} = 8.8

 

 

 

 

 


Option 1)

+11.2

Option 2)

-11.2

Option 3)

- 8.8

Option 4)

+ 8.8

Posted by

Avinash

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