Question

Derive the expression for refractive index of the material of the prism in terms of angle of the prism and angle of minimum deviation.

Answer 1

In the given diagram,

$OP$ is the incidence ray, which is making the angle $i_1$ with normal, and $QR$ is the angle of emergence, which is represented by $i_2$. $A$ is the prism angle and $\mu$ is the refractive index of the prism.

Now, We know that,

$A=$Prism angle, $\delta =$Angle of deviation, $i_1=$Angle of incidence, $i_2=$Angle of emergent.

In the case of minimum deviation,$\angle r_1=\angle r_2=\angle r$

$A=\angle r_1+\angle r_2$

SO, $A=\angle r+\angle r=\angle 2r$

$\angle r=\frac{A}{2}$

Now, again

$A+\delta =i_1+i_2(\because$ In the case of minimum deviation

$i_1=i_2=i$ and $\delta ={\delta }_m)$

So, $A+{\delta }_m=i+i=2i$

Now, $i=\frac{A+{\delta }_m}{2}$

Now, from snell's rule,

$\mu =\frac{sini}{sinr}$

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