Q163 / 170 · Manipal (MET) 2020 · Physics

The magnifying power of a telescope is $m^{\prime}$. If the focal length of the eye piece is doubled, then its magnifying power will become

  • A
    $\sqrt{2} \mathrm{~m}$
  • B
    $3 m$
  • C
    $2 m$
  • Correct answer
    D
    $\frac{m}{2}$

Solution

Magnifying power of a telescope is given by $m=\frac{f_o}{f_e}$ If the focal length of the eye piece is doubled, the new magnifying power is given by $m^{\prime}=\frac{f_o}{f_e^{\prime}}=\frac{f_o}{2 f_e}$ $\therefore \quad m^{\prime}=\frac{m}{2}$

Hence, the correct answer is option: 'D'.

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