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: C
Solution :
Using the relation for the focal length of concave mirror \[\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\] ?(1) Differentiating equation (1), we obtain \[0=-\frac{1}{{{\upsilon }^{2}}}dv-\frac{1}{{{u}^{2}}}du\] So, \[dv=-\frac{{{\upsilon }^{2}}}{{{u}^{2}}}\times b\] ?(2) (Here :\[du=b\]) From equation (1) \[\frac{1}{\upsilon }=\frac{1}{f}-\frac{1}{u}=\frac{u-f}{fu}\] or \[\frac{u}{\upsilon }=\frac{u-f}{f}\] \[\frac{v}{u}=\frac{f}{u-f}\] ?(3) Now, from equations (2) \[d\upsilon =-{{\left( \frac{f}{u-f} \right)}^{2}}b\] 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
