Angle of deviation ($\delta$) is the angle between emergent ray and incident ray. 
For a single refracting surface, $\delta =|i-r|$
For a prism, $\delta =(i_1+i_2)-(r_1+r_2)$ 
                      $\delta =i_1+i_2-A$ where $A$ is the angle of the prism.
For angle of minimum deviation, $\delta$ is minimum and $i_1=i_2=i$
                      ${\delta }_{min}=2i-A$
For small $A$, $\delta =(\mu -1)A$

Example: If the angle of minimum deviation produced by an equilateral prism is 30${}^{\circ }$ , then find the refractive index of material of prism.

Solution:
given,      $A=60^{\circ }$

$\delta m=30^{\circ }$

$\mu =\frac{sin\frac{\delta m+A}{2}}{sin\frac{A}{2}}$

$\mu =\frac{sin\frac{60+30}{2}}{sin\frac{60}{2}}$

=$\sqrt{2}$

In determining the limit of resolution of optical instruments like a telescope or a microscope, for the two stars to be just resolved,
$f\Delta \theta \approx r_0\approx \frac{0.61\lambda f}{a}$
implying $\Delta \theta \approx \frac{0.61\lambda }{a}$
Thus $\Delta \theta$ will be small if the diameter of the objective is large. This implies that the telescope will have better resolving power if $a$ is large. It is for this reason that for better resolution, a telescope must have a large diameter objective.