$(a)\;\sqrt{\large\frac{\lambda R}{\lambda R-1}}\qquad(b)\;\sqrt{\large\frac{(\lambda R-1)}{\lambda R}}\qquad(c)\;\sqrt{\lambda(R-1)}\qquad(d)\;\sqrt{\large\frac{\lambda R}{\lambda R-1}}$

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Answer : (d) $\;\sqrt{\large\frac{\lambda R}{\lambda R-1}}$

Explanation :

As $\; \large\frac{1}{\lambda}=R\;(\large\frac{1}{n_{1}^2}-\large\frac{1}{n_{2}^2})$

$\large\frac{1}{\lambda}=R\;(\large\frac{1}{1^2}-\large\frac{1}{n^2})$

$\large\frac{1}{\lambda R}=1-\large\frac{1}{n^2} \quad \; or \qquad \large\frac{1}{n^2}=1-\large\frac{1}{\lambda R}=\large\frac{\lambda R-1}{\lambda R}$

$n=\sqrt{\large\frac{\lambda R}{\lambda R-1}}\;.$

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