A) \[\left( \frac{f}{u-f} \right)b\]
B) \[{{\left( \frac{f}{u-f} \right)}^{2}}b\]
C) \[\left( \frac{f}{u-f} \right){{b}^{2}}\]
D) \[\left( \frac{f}{u-f} \right)\]
Correct Answer: B
Solution :
Using the relation for the focal length of concave mirror \[\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\] ... (i) Differentiating Eq. (i), we obtain \[0=-\frac{1}{{{v}^{2}}}dv-\frac{1}{{{u}^{2}}}du\] So \[av=-\frac{{{v}^{2}}}{{{u}^{2}}}\times b\] ... (ii) (Here,\[du=b\]) From Eq. (i) \[\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{u-f}{f\,\,v}\] or \[\frac{u}{v}=\frac{u-f}{v}\] \[\frac{v}{u}=\frac{f}{u-f}\] ? (ii) Now, from Eqs. (ii) and (iii), we get \[dv=-{{\left( \frac{f}{u-f} \right)}^{2}}b\] Therefore, size of image is\[={{\left( \frac{f}{u-f} \right)}^{2}}b\]
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