Which of the following graphs correctly represents the variation of angle of deviation $\left(\delta \right)$ versus angle of incidence $\left(i\right)$ for a light ray travelling from a denser to rarer medium?

The correct option is B
$\text{Case I}:i<C$
If angle of incidence lesser than the critical angle, $\text{i.e}$ $i<C$, refraction occurs.

Angle of deviation, $\delta$$=r-i.....\left(1\right)$

According to the Snell'e law
${\mu }_d\sin i={\mu }_r\sin r$

$\Rightarrow r={\sin }^{-1}\left(\frac{{\mu }_d}{{\mu }_r}\sin i\right)$

Putting the value of $r$ in the equation $\left(1\right)$, we get
$\Rightarrow \delta ={\sin }^{-1}\left(\frac{{\mu }_d}{{\mu }_r}\sin i\right)-i.....\left(2\right)$

When,
$i=0^{\circ }$ from $\left(2\right)$,

$\Rightarrow \delta ={\sin }^{-1}\left(\frac{{\mu }_d}{{\mu }_r}\sin 0^{\circ }\right)-0^{\circ }$

$\therefore \delta =0$

When,
$i=C$, from $\left(2\right)$,

$\Rightarrow \delta ={\sin }^{-1}\left(\frac{{\mu }_d}{{\mu }_r}\sin C\right)-C$

As we know, $(\sin C=\frac{{\mu }_r}{{\mu }_d})$

$\Rightarrow \delta ={\sin }^{-1}\left(\frac{\sin C}{\sin C}\right)-C$

$\therefore \delta =\frac{\pi }{2}-C........\left(3\right)$

$\text{Case II}:i>C$

From ray diagram, angle of deviation, $\delta =\pi -2i$
when, $i=C$, we have
$\delta =\pi -2C$
Comparing above equation with the standard straight line equation,
$y=-mx+c$
We can write, $\delta =-2C+\pi ........\left(4\right)$
There is a discontinuity at $i=C$.

When equation $\left(2\right)$ and $\left(4\right)$ are plotted , we have graph as shown below.

Why this Question : This question checks the variation of angle of deviation $\left(\delta \right)$ verses $i$ for a prism under all possible circumstances.