A) \[\left( \frac{f}{u-f} \right)b\]
B) \[{{\left( \frac{f}{u-f} \right)}^{2}}b\]
C) \[\left( \frac{f}{u-f} \right)b2\]
D) \[\left( \frac{f}{u-f} \right)\]
Correct Answer: B
Solution :
Using the relation for the focal length of concave mirror \[\frac{1}{f}=\frac{1}{\upsilon }+\frac{1}{u}\] ...(i) Differentiating equation (i), we obtain \[0=-\frac{1}{{{\upsilon }^{2}}}d\upsilon -\frac{1}{{{u}^{2}}}du\] So, \[d\upsilon =-\frac{{{\upsilon }^{2}}}{{{u}^{2}}}\times b\] ...(ii) (Here:\[d\upsilon =b\]) From equation (i) \[\frac{1}{\upsilon }=\frac{1}{f}-\frac{1}{u}=\frac{u-f}{fu}\] Or \[\frac{u}{\upsilon }=\frac{u-f}{f}\] \[\frac{\upsilon }{u}=\frac{f}{u-f}\] ?..(iii) Now, from equations (ii) and (iii), we get \[d\upsilon =-{{\left( \frac{f}{u-f} \right)}^{2}}b\] 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
