In an astronomical telescope in normal adjustment,a straight black line of length $L$ is drawn on the objective lens. The eyepiece forms a real image of this line. The length of this image is $l$. The magnification of the telescope is

  • A
    $L/l$
  • B
    $L/l + 1$
  • C
    $L/l - 1$
  • D
    $(L + l)/(L - l)$

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Similar Questions

The focal length of the objective and the eyepiece of a telescope are $100\, cm$ and $5\, cm$ respectively. If the final image is formed at the least distance of distinct vision,the magnification of the telescope is:

$A$ telescope has an objective of focal length $50 \text{ cm}$ and an eyepiece of focal length $5 \text{ cm}$. The least distance of distinct vision is $25 \text{ cm}$. The telescope is focused for distinct vision on a scale $200 \text{ cm}$ away. The separation between the objective and the eyepiece is.......$\text{cm}$.

All of the following statements are correct except:

For a telescope in normal adjustment,the length of the telescope is found to be $27 \, cm$. If the magnifying power of the telescope at normal adjustment is $8$,the focal lengths of the objective and the eyepiece are respectively:

An astronomical refracting telescope will have large angular magnification and high angular resolution when it has an objective lens of

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