A) \[L{{\left[ \frac{f}{u-f} \right]}^{1/2}}\]
B) \[L{{\left[ \frac{(u+f)}{f} \right]}^{1/2}}\]
C) \[L{{\left[ \frac{f}{(u-f)} \right]}^{2}}\]
D) \[L{{\left[ \frac{f}{(u+f)} \right]}^{2}}\]
Correct Answer: C
Solution :
\[\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\]or \[-\frac{dv}{{{v}^{2}}}-\frac{du}{{{u}^{2}}}=0\]i.e., \[dv=-du{{(v/u)}^{2}}\] But \[v=\frac{uf}{u-f}\]or \[\frac{v}{\upsilon }=\frac{f}{u-f}\]so, \[dv=-du{{\left\{ \frac{f}{u-f} \right\}}^{2}}\] or \[\left| dv \right|=L{{\left\{ \frac{f}{u-f} \right\}}^{2}}\] Hence, the correction option is [c].
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