Q87 / 100 · BITSAT 2024 (Memory Based Paper 2) · Physics

In normal adjustment, for a refracting telescope, the distance between objective and eye piece is 30 cm . The focal length of the objective, when the angular magnification of the telescope is 2 , will be:

  • Correct answer
    A
    20 cm
  • B
    30 cm
  • C
    10 cm
  • D
    15 cm

Solution

$\because \mathrm{f}_0+\mathrm{f}_{\mathrm{e}}=30$ And magnification, $\mathrm{m}=\frac{\mathrm{f}_0}{\mathrm{f}_e}$ $2=\frac{\mathrm{f}_0}{\mathrm{f}_{\mathrm{e}}} \Rightarrow \mathrm{f}_0=2 \mathrm{f}_{\mathrm{e}} \Rightarrow \mathrm{f}_0+\frac{\mathrm{f}_0}{2}=30 \therefore \mathrm{f}_0=20cm$

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

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