Answer:
Option A
Explanation:
As we know, v= n$\lambda$
$\Rightarrow $ $\frac{1}{\lambda}=\frac{n}{v}\Rightarrow \frac{1}{\lambda}=R\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)$
$\Rightarrow $ $v=Rc\left(\frac{1}{n^{2}_{1}}-\frac{1}{n^{2}_{2}}\right)$
$\therefore$
$v_{2}=Rc\left(\frac{1}{2^{2}}-\frac{1}{3^2}\right)=Rc\left(\frac{1}{4}-\frac{1}{9}\right)$ .....(i)
$v_{1}=Rc\left(\frac{1}{2^{2}}\right)=\frac{Rc}{4}$
$v_{3}=Rc\left(\frac{1}{3^{2}}\right)=\frac{Rc}{9}$
$\Rightarrow$ $v_{1}-v_{3}=Rc\left(\frac{1}{4}-\frac{1}{9}\right)$ ......(ii)
From Eqs.(i) and (ii), we get
v1 -v3 =v2
$\Rightarrow$ $v_{1}-v_{2}=v_{3}$