Q135 / 160 · AP EAMCET 2010 · Physics

The diameter of objective of a telescope is $1 \mathrm{~m}$ Its resolving limit for the light of wavelength $4538 Ã…$, will be

  • Correct answer
    A
    $5.54 \times 10^{-7} \mathrm{rad}$
  • B
    $2.54 \times 10^{-4} \mathrm{rad}$
  • C
    $6.54 \times 10^{-7} \mathrm{rad}$
  • D
    None of these

Solution

Resolving limit $d \theta=\frac{1.22 \lambda}{a}$ $=\frac{1.22 \times 4538 \times 10^{-10}}{1}$ $=5.54 \times 10^{-7} \mathrm{rad}$

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

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