Why do we have minimum deviation angle in a Prism? why do we have minimum deviation angle in a prism? I mean why does the angle of emergence first decreases, attains a minimum value and then increases.why?

 A: I would like to answer this question step by step. Let's start from angle of deviation.

Why do we need a measure called angle of deviation? 

It's because a ray of light undergo deviation from its actual path if it is allowed to pass from one medium to a different medium. This is due to the change in index of refraction of the media. The angle of deviation is a measure of how much this deviation has happened at the interface. In the case of a prism (usually made of glass), light enters the prism from air. Since air and prism have different refractive indices, light undergo a deviation from it's actual path. Let's call this deviation at the air-glass interface the angle of deviation.  This is as illustrated in the figure below. The dotted line represents the undeviated direction of the ray if the prism was no there.

Picture source: $\small{\text{https://en.wikipedia.org/wiki/File:Deflection_of_light_ray_passing_through_prism.png}}$

What is the angle of minimum deviation? 

Of course, this is the minimum value of angle of deviation. At the minimum deviation angle, the angle of incidence and  the angle of emergence of the ray are identical. As you can see from the above figure, this will occur only if the light ray inside the prism passes parallel to the base of the prism (you can prove it from the prism equation. See for example: https://www.physics.wustl.edu/ClassInfo/316/Theory/Refraction.pdf). Also, this angle of deviation happens only for a particular angle of incidence.   

$\small{\text{Picture source: https://en.wikipedia.org/wiki/File:Minimum_deflection_of_light_ray_passing_through_prism.png}}$    
If the angle of prism is $A$, the deviation angle for a particular angle of incidence $i$ be $D$ and the corresponding angle of emergence be $e$. Then we have the relation, 
$$i+e=D+A$$  
At $D=D_{min}$, then $i=e$, which implies that  
$$D_{min}=2i-A$$

Why does the angle of emergence first decreases, attains a minimum value and then increases?  

When a ray of light is incident obliquely on the interface of another medium,  refraction happens. However, if the ray of light is incident normally on the interface, then refraction will not happen (actually the refraction is always there and in this particular scenario, the angle of refraction becomes zero), and the the ray of light passes undeviated through the prism, leaving the angle of deviation zero.  This is the normal position.  
When you rotate the prism about an axis passing perpendicular to the plane of incidence, then you can see that the angle of deviation changes with change in angle of incidence and finally reaches a minimum value and after that it increases again on further rotation in the same direction.  
If the incident light beam is rotated in either direction from the minimum deviation position, the deviation of the light from its incident path caused by refraction in the prism will be greater. So the angle of refraction emergence will be greater as you move either way from the minimum deviation position. This means that if you move along a particular direction, then the angle of emergence first decreases (on rotating towards the angle of minimum deviation position), then reaches a minimum value (equal to the angle of incidence) at the minimum deviation position and after that it again starts to increase with increase in the angle of incidence.
A: I attempt to explain the presence of a minimum mathematically instead of experimentally.
The light ray travels through air and hits the glass prism, at an angle $\theta_1$ (the angle of incidence) to the normal of the glass boundary.
When the ray is in the glass, the angle of the ray changes to $\theta_2$ (the angle of refraction), to the normal of the glass boundary. The two angles are related by Snell's Law, such that:
$$\frac{sin\theta_1}{sin\theta_2} = \frac{n_{glass}}{n_{air}}$$
where $n_{air}$ is the refractive index of air, and $n_{glass}$ is the refractive index of glass. Since $n_{glass} > n_{air}$, $\theta_2 < \theta_1$.
Assuming you know $\theta_1$, $n_{glass}$, and $n_{air}$, then you can find out $\theta_2$.
Then, when the light ray (which is currently inside the glass) hits the air boundary, there is a new angle of incidence $\theta_3$. $\theta_3$ depends on the the angle at the top of the prism, $A$.
$$sum\space of\space angles\space in\space triangle = 180^{\circ}$$
$$180^{\circ} = A + (90^{\circ} - \theta_3) + (90^{\circ} - \theta_2)$$
$$\theta_3 = A - \theta_2$$
We want to know the angle of refraction $\theta_4$. So we apply Snell's Law again:
$$\frac{sin\theta_4}{sin\theta_3} = \frac{n_{glass}}{n_{air}}$$

From geometry,
$$angle\space of\space deviation = \theta_1-\theta_2+\theta_4-\theta_3$$

Substituting $\theta_3 = A - \theta_2$,
$$angle\space of\space deviation = \theta_1+\theta_4-A$$
To minimize $angle\space of\space deviation$, assuming A is constant, we need to minimize $\theta_1+\theta_4$.
We can find $\theta_4$ in terms of $\theta_1$. Equating the two Snell's Law equations,
$$\frac{sin\theta_4}{sin\theta_3} = \frac{sin\theta_1}{sin\theta_2}$$
Substituting $\theta_3 = A - \theta_2$,
$$\frac{sin\theta_4}{sin(A - \theta_2)} = \frac{sin\theta_1}{sin\theta_2}$$
$$sin\theta_4 = (sinA\space cos\theta_2 - cosA\space sin\theta_2) * \frac{sin\theta_1}{sin\theta_2}$$
$$sin\theta_4 = (\frac{sinA}{tan\theta_2} - cosA) * sin\theta_1$$
Substituting $\theta_2 = sin^{-1}(\frac{n_{air}*sin\theta_1}{n_{glass}})$,
$$\theta_4 = sin^{-1}\Bigg(\bigg(\frac{sinA}{tan(sin^{-1}(\frac{n_{air}*sin\theta_1}{n_{glass}}))} - cosA\bigg) * sin\theta_1\Bigg)$$
Because I'm not good at math, this is where my analysis ends, and WolframAlpha begins. Punching in arbitrary values for the angle of prism, and for refractive indeces, we find the values of $\theta_1$ such that we minimize $\theta_1 + \theta_4$.
The very fact that WolframAlpha can find minima for $\theta_1 + \theta_4$ means that there are values of $\theta_1$ such that $angle\space of\space deviation$ has minima.
Notice that the values of the minimum are offset from the value of $\theta$ by a constant amount every time. This constant amount happens to be $\theta_1$ such that $\theta_1 = \theta_4$. So, this shows that the $angle\space of\space deviation$ is minimum only for $\theta_1 = \theta_4$.
