$(a)\;l=\large\frac{2 n \lambda}{\sqrt{3}-1}\qquad(b)\;l=\large\frac{(2 n-1) \lambda}{2(\sqrt{3}-1)}\qquad(c)\;l=\large\frac{(2 n-1) \lambda \sqrt{3}}{4(2-\sqrt{3}}\qquad(d)\;l=\large\frac{(2 n-1) \lambda}{\sqrt{3}-1}$

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