Question
Draw the graph of deviation angle $(\delta)$ versus incidence angle (i) and derive the formula for refractive index $n_{21}=\frac{\sin \left(\frac{A+D_m}{2}\right)}{\sin \left(\frac{A}{2}\right)}$ for the material of prism.
✓
Answer
→The graph of deviation angle versus incidence angle is shown in figure.
→The graph shows that for a single value of deviation angle ( $\delta$ ) there are two values of incidence angle $i$ and hence also of $e$.
→From the symmetry it can be said that angle of deviation $\delta$ remains the same if angle of incidence $i$ and angle of emergent $e$ are interchanged. Even if the path of ray can be traced back, resulting in the same angle of deviation.
→From the graph, for a particular value of $i=e$ the angle of incidence, a single value of deviation is obtained. At the minimum deviation, $D _m$, the refracted ray becomes parallel to its base.
→So when $\delta= D _m$ and $i=e$, then $r_1=r_2$.
→For prism, $\begin{aligned} A & =r_1+r_2 \\ \therefore \quad A & =2 r_1 \\ \therefore \quad r_1 & =\frac{ A }{2}......(1)\end{aligned}$
→and from $\delta=i+e- A$,
$\begin{array}{l} D _m=2 i- A \\2 i= D _m+ A \\i=\frac{ A + D _m}{2}......(2)\end{array}$
→Applying Snell's law at incident point Q ,
$n_1 \sin i=n_2 \sin r_1$
→Substituting value of $r_1$ and $i$ from equation (1) and (2),
$\begin{array}{l}\therefore n_1 \sin \left(\frac{ A + D _m}{2}\right)=n_2 \sin \left(\frac{ A }{2}\right) \\\therefore \quad \frac{n_2}{n_1}=\frac{\sin \left(\frac{ A + D _m}{2}\right)}{\sin \frac{ A }{2}} \\\therefore \quad n_{21}=\frac{\sin \left(\frac{ A + D _m}{2}\right)}{\sin \frac{ A }{2}}\end{array}$
→which is the formula to find the refractive index of the material of the prism.
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