Question

The diameter and the focal length of the objective of an astronomical telescope are $140$ mm and $3.5$ m respectively & focal length of the eyepiece is 25 mm. Find the magnifying power and resolving power of this telescope. Given : wavelength of yellow colour = $6000A^{\ast }$.

Answer 1

$l=l_{\circ }\left(1+\alpha \theta \right)$

$\Rightarrow l=l_{\circ }+\mu \alpha \theta$

$\Rightarrow \Delta l=l_{\circ }\alpha \theta$

also

$\gamma =\frac{F{l}_{\circ }}{A\Delta l}$

$\Rightarrow F=\gamma A\frac{\Delta l}{l_{\circ }}=\gamma \pi r^2\alpha \theta$

$\therefore F=\pi r^2\gamma \alpha \theta$