Question

A telescope is used to resolve two stars separated by $4.6\times {10}^{-6}$ rad. If the wavelength of light used is 5460 $\stackrel{0}{A}$, what should be the aperture of the objective of the telescope?

• A. 0.1488 m
• B. 0.567 m
• C. 1 m
• D. 2 m

Answer 1

The correct option is A 0.1488 m

Aperature of the telescope, $D=\frac{1.22\lambda }{d\theta }$

Here, $\lambda =5460\stackrel{0}{A}=5460\times {10}^{-10}m$

$d\theta =4.6\times {10}^{-6}rad$

$\therefore D=\frac{1.22\times 5460\times {10}^{-10}}{4.6\times {10}^{-6}}$

$=0.1488m$