A) \[l{{\left( \frac{u-f}{f} \right)}^{1/2}}\]
B) \[l{{\left( \frac{u-f}{f} \right)}^{2}}\]
C) \[l{{\left( \frac{f}{u-f} \right)}^{1/2}}\]
D) \[l{{\left( \frac{f}{u-f} \right)}^{2}}\]
Correct Answer: D
Solution :
From mirror formula \[\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\] .....(i) Differentiating equation (i), we obtain \[0=-\frac{1}{{{v}^{2}}}dv-\frac{1}{{{u}^{2}}}du\] \[\Rightarrow dv=-{{\left( \frac{v}{u} \right)}^{2}}du\] .....(ii) Also from equation (i) \[\frac{v}{u}=\frac{f}{u-f}\] .....(iii) From equation (ii) and (iii) we get \[dv=-{{\left( \frac{f}{u-f} \right)}^{2}}.\ l\] Therefore size of image is \[{{\left( \frac{f}{u-f} \right)}^{2}}l.\]
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