Question
An astronomical telescope is an optical instrument which is used for observing distinct images of heavenly bodies libe stars, planets etc. It consists of two lenses. In normal adjustment of telescope, the final image is formed at infinity. Magnifying power of an astronomical telescope in normal adjustment is defined as the ratio of the angle subtended at the eye by the angle subtended at the eye by the final image to the angle subtended at the eye, by the object directly, when the final image and the object both lie at infinite distance from the eye. It is given by, $\text{m}=\frac{\text{f}_0}{\text{f}_\text{g}}.$ To increase magnifying power of an astronomical telescope in normal adjustment, focal length of objective lens should be large and focal length of eye lens should be small.
  1. An astronomical telescope of magnifying power 7 consists of the two thin lenses 40cm apart, in normal adjustment. The focal lengths of the lenses are
  1. 5cm, 35cm
  2. 7cm, 35cm
  3. 17cm, 35cm
  4. 5cm, 30cm
  1. An astronomical telescope has a magnifying power of 10. In normal adjustment, distance between the objective and eye piece is 22cm. The focal length of objective lens is:
  1. 25cm
  2. 10cm
  3. 15cm
  4. 20cm
  1. In astronomical telescope compare to eye piece, objective lens has:
  1. Negative focal length.
  2. Zero focal length.
  3. Small focal length.
  4. Large focal length.
  1. To see stars, use:
  1. Simple microscope.
  2. Compound microscope.
  3. Endoscope.
  4. Astronomical telescope.
  1. For large magnifying power of astronomical telescope.
  1. f0 << fe
  2. f0 << fe
  3. f0 << fe
  4. None of these.

Answer

  1. (a) 5cm, 35cm

Explanation:

$\text{m}=\frac{\text{f}_0}{\text{f}_\text{e}}=7$

$\text{f}_0=7\text{f}_e$

In normal adjustment, distance between the lenses,

$\text{f}_0+\text{f}_\text{e}=40$

$7\text{f}_0+\text{f}_\text{e}=40\Rightarrow\text{f}_\text{e}=\frac{40}{8}=5\text{cm}$

$\text{f}_0=7\text{f}_\text{e}=7\times5=35\text{cm}$

  1.  (d) 20cm

Explanation:

$\text{m}=-10;\text{L}=22\text{cm}$

As, $\text{m}=\frac{-\text{f}_0}{\text{f}_e}\Rightarrow-10=-\frac{\text{f}_0}{\text{f}_\text{e}}$

$\text{f}_0=10\text{f}_\text{e}$

As, $\text{L}=\text{f}_0+\text{f}_\text{e}$

$22=10\text{f}_\text{e}+\text{f}_e=11\text{f}_\text{e}$

or $\text{f}_\text{e}=\frac{22}{11}=2\text{cm}$

$\text{f}_0=10\text{f}_\text{e}=20\text{cm}$

  1.  (d) Large focal length.

Explanation:

Objective lens has larger focal length than eye-piece.

  1.  (d) Astronomical telescope.

Explanation:

Astronomial telescope is used to see stars, sun etc.

  1. (c) f0 << fe

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