Question

A telescope can just resolve $($Rayleigh Criterion$)$ two objects which are $2000\text{km}$ away from it. If the separation between objects is $10\text{m}$ and the wavelength of light used is $500\text{nm}$ then the minimum diameter of the objective of the telescope is

• A. $12.2\text{cm}$
• B. $24.4\text{cm}$
• C. $30.0\text{cm}$
• D. $48.8\text{cm}$

Answer 1

The correct option is A $12.2\text{cm}$


$\sin \theta =\frac{1.22\lambda }{D}$
Since, $\theta$ is very small, so, $\theta =\frac{1.22\lambda }{D}$
Also, $\theta =\frac{10}{2000\times {10}^3}$

Further,
$\frac{10}{2000\times {10}^3}=\frac{1.22\lambda }{D}$ $\Rightarrow D=1.22\times 500\times {10}^{-9}\times 200\times {10}^3$
$\Rightarrow D=0.122\text{m}=12.2\text{cm}$