A) \[L{{\left[ \frac{f}{u-f} \right]}^{1/2}}\]
B) \[L{{\left[ \frac{u+f}{f} \right]}^{1/2}}\]
C) \[L{{\left[ \frac{u-f}{f} \right]}^{2}}\]
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 Eq. (i), we obtain \[0=\frac{1}{{{v}^{2}}}dv-\frac{1}{{{v}^{2}}}du\] \[dv=-{{\left( \frac{v}{u} \right)}^{2}}du\] ?. (ii) Also from Eq. (i) \[\frac{v}{u}=\frac{f}{u-f}\] ... (iii) From Eqs. (ii) and (iii), we get \[dv=-{{\left( \frac{f}{u-f} \right)}^{2}}\] Therefore size of image is \[{{\left( \frac{f}{u-f} \right)}^{2}}L\]
You need to login to perform this action.
You will be redirected in
3 sec
