A) \[\left( \frac{f}{u-f} \right)b\]
B) \[{{\left( \frac{f}{u-f} \right)}^{2}}b\]
C) \[{{\left( \frac{f}{u-f} \right)}^{2}}{{b}^{2}}\]
D) \[\left( \frac{f}{u-f} \right){{b}^{2}}\]
Correct Answer: B
Solution :
From relation the focal length of concave mirror is given by: \[\frac{1}{f}=\frac{1}{\upsilon }+\frac{1}{u}\] ?..(i) Differentiating equation (i), we obtain, \[0=-\frac{1}{\upsilon }d\upsilon -\frac{1}{{{u}^{2}}du.}\] or \[d\upsilon =-\frac{{{\upsilon }^{2}}}{{{u}^{2}}}du\] Here \[du=b\] \[d\upsilon =-\frac{{{\upsilon }^{2}}}{{{u}^{2}}}\times b\] ?..(ii) From equation (i) we have, \[\frac{1}{\upsilon }=\frac{1}{f}-\frac{1}{u}=\frac{u-f}{fu}\] \[\Rightarrow \] \[\frac{u}{\upsilon }=\frac{u-f}{f}\] \[\Rightarrow \] \[\frac{u}{\upsilon }=\frac{f}{u-f}\] ??(iii) Finally from equation (ii) and (iii) we get \[d\upsilon =-{{\left( \frac{f}{u-f} \right)}^{2}}b\] Therefore size of image \[={{\left( \frac{f}{u-f} \right)}^{2}}b.\]
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