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.

A plano convex lens of focal length $30cm$ has its plane surface silvered. An object is placed $40cm$ from the lens on the convex side. The distance of the image from the lens is:
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
or, $\frac{1}{v}=\frac{1}{30}-\frac{1}{40}$
or, $\frac{1}{v}=\frac{4-3}{120}$
or, $v=+120cm$
Now, this image acts as the object for the lens.
So, $u=+120cm,f=30cm$
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
or, $\frac{1}{v}=\frac{1}{30}+\frac{1}{120}$
or, $v=+24cm$

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 $A$, $\delta =(\mu -1)A$

Problem:
A convex lens, of focal length $30cm$, a concave lens of focal length $120cm$, and a plane mirror are arranged as shown. For an object kept at a distance of $60cm$ from the convex lens, the final image, formed by the combination, is a real image, at a distance of :
Solution:
Location of image formation after first refraction from the lens,$\frac{1}{v}+\frac{1}{60}=\frac{1}{30}$.
This gives $v=60cm$
Next refraction from concave lens, $\frac{1}{v_2}-\frac{1}{40}=-\frac{1}{120}$
Hence $v_2=60cm$
Since this will be formed behind the mirror, this will be final image.It is at a distance of $60cm$ from the concave lens.

Real Depth is actual distance of an object beneath the surface, as would be measured by submerging a perfect ruler along with it.

Apparent depth in a medium is the depth of an object in a denser medium as seen from the rarer medium. Its value is smaller than the real depth.
$D_{apparent}=\frac{D_{real}}{\mu }$

When a ray of light enters from a denser medium to a rarer medium,it bends away from the normal, for example, the ray $A{O}_{1}B$ .The incident ray $A{O}_1$ is partially reflected ($O_{1}C$) and partially transmitted($O_{1}B$) or refracted, the angle of refraction (r ) being larger than the angle of incidence (i ). As the angle of incidence increases, so does the angle of refraction, till for the ray $A{O}_3$, the angle of refraction is $\frac{\pi }{2}$.The refracted ray is bent so much away from the normal that it grazes the surface at the interface between the two media. This is shown by the ray $A{O}_{3}D$ If the angle of incidence is increased still further (e.g.,the ray $A{O}_4$ ), refraction is not possible, and the incident ray is totally reflected.

Example: At what angle should a ray of light be incident on the face of a prism of refracting angle ${60}^{\circ }$ so that it just suffers total internal reflection at the other face. Find the refractive index of the prism.Refractive index of prism is 1.524.

Solution:
$$\begin{gathered} \sin c=\frac{1}{\mu }=\frac{1}{1.524}=0.65orc={40}^{\circ }{30}^{{}^{\prime }} \\ r={19}^{\circ }{30}^{{}^{\prime }}=\left[{180}^{\circ }{-120}^{\circ }{-40}^{\circ }{30}^{{}^{\prime }}\right] \\ \mu =\frac{\sin i}{\sin r}or\sin i=1.524sin{19}^{\circ }{30}^{{}^{\prime }} \\ =1.524\left(.3340\right)=0.5 \\ ori={30}^{\circ } \end{gathered}$$

Suppose while sitting in a parked car, you notice a jogger approaching towards you in the side view mirror of $R=2m$. If the jogger is running at a speed of $5m{s}^{-1}$, how fast the image of the jogger appear to move when the jogger is $39m$ away?
Solution:
From the mirror equation, we get:
$v=\frac{uf}{u-f}$
For convex mirror, since $R=2m,f=1m$, then for $u=39m$, $v=\frac{-39\times 1}{-39-1}$
Since the jogger moves at a constant speed of $5m{s}^{-1}$, after $1s$, the position of the image $v$ (for $u=39-5=34$) is $\frac{34}{35}m$.
The shift in the position of image in $1s$ is
$\frac{39}{40}-\frac{34}{35}=\frac{1}{280}m$
Therefore, the average speed of the image when the jogger is between $39m$ and $34m$ from the mirror, is $\frac{1}{280}m{s}^{-1}$.

Consider a convex lens (or concave lens) of absolute refractive index ${\mu }_2$ to be placed in a rarer medium of absolute refractive index ${\mu }_1$. Considering the refraction of a point object on the surface $X{P}_{1}Y$, the image is formed at $I_1$ who is at a distance of $V_1$.

$C{I}_1=P_1{I}_1=V_1$ (as the lens is thin)

$C{C}_1=P_1{C}_1=R_1$

$CO=P_{1}O=u$

It follows from the refraction due to convex spherical surface $X{P}_{1}Y$

$\frac{{\mu }_1}{-u}+\frac{{\mu }_2}{v_1}=\frac{{\mu }_2-{\mu }_1}{R_1}..........(i)$

The refracted ray from A suffers a second refraction on the surface $X{P}_{2}Y$ and emerges along BI. Therefore I is the final real image of O.

Here the object distance is

$u=C{I}_1\approx P_2{I}_1=V$

$P_1{P}_2$ is very small

Let $CI\approx P_{2}I=V$

(Final image distance)

Let $R_2$ be radius of curvature of second surface of the lens. It follows from refraction due to concave spherical surface from denser to rarer medium that

$\frac{-{\mu }_2}{v_1}+\frac{{\mu }_1}{v}=\frac{{\mu }_1-{\mu }_2}{R_2}=\frac{{\mu }_2-{\mu }_1}{-R_2}............(ii)$

Adding $(i)and(ii)$

$\frac{-{\mu }_1}{u}+\frac{{\mu }_1}{v}=({\mu }_2-{\mu }_1)(\frac{1}{R_1}-\frac{1}{R_2})$

or

${\mu }_1(\frac{1}{v}-\frac{1}{u})=({\mu }_2-{\mu }_1)(\frac{1}{R_1}-\frac{1}{R_2})$

But $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ and $\frac{{\mu }_2}{{\mu }_1}=\mu$

Thus we get=

$\frac{1}{f}=(\mu -1)(\frac{1}{R_1}-\frac{1}{R_2})$

Problem:
A convex lens, of focal length $30cm$, a concave lens of focal length $120cm$, and a plane mirror are arranged as shown. For an object kept at a distance of $60cm$ from the convex lens, the final image, formed by the combination, is a real image, at a distance of :
Solution:
Location of image formation after first refraction from the lens,$\frac{1}{v}+\frac{1}{60}=\frac{1}{30}$.
This gives $v=60cm$
Next refraction from concave lens, $\frac{1}{v_2}-\frac{1}{40}=-\frac{1}{120}$
Hence $v_2=60cm$
Since this will be formed behind the mirror, this will be final image.It is at a distance of $60cm$ from the concave lens.