(A) The resolving limit $(d\theta)$ of a telescope is given by the formula:$d\theta = \frac{1.22 \lambda}{a}$Where $\lambda$ is the wavelength of ligh

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(A) The resolving limit $(d\theta)$ of a telescope is given by the formula:$d\theta = \frac{1.22 \lambda}{a}$Where $\lambda$ is the wavelength of light and $a$ is the diameter of the objective lens.Given:$\lambda = 4538 \ \mathring A = 4538 \times 10^{-10} \ m$$a = 1 \ m$Substituting the values:$d\theta = \frac{1.22 \times 4538 \times 10^{-10}}{1}$$d\theta = 5536.36 \times 10^{-10} \ rad$$d\theta \approx 5.54 \times 10^{-7} \ rad$Thus,the correct option is $A$.

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The diameter of the objective of a telescope is $1 \ m$. Its resolving limit for light of wavelength $4538 \ \mathring A$ will be:

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

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The diameter of the objective of a telescope is $1 \ m$. Its resolving limit for light of wavelength $4538 \ \mathring A$ will be:

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