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 \[dv=-\frac{{{v}^{2}}}{{{u}^{2}}}\times b\] ... (ii) (Here:\[du=b\]) From Eq. (i) \[\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{u-f}{fu}\] or \[\frac{u}{v}=\frac{u-f}{f}\] \[\frac{v}{u}=\frac{f}{u-f}\] ... (ii) Now, from Eqs. (ii) and (iii), we get \[dv=-{{\left( \frac{f}{u-f} \right)}^{2}}b\] Therefore, size of image is\[={{\left( \frac{f}{u-f} \right)}^{2}}b\]
You need to login to perform this action.
You will be redirected in
3 sec
