A short linear object of length L lies on the axis of a spherical mirror of focal length f at a distance b from the mirror. The size of the image is:
Answer & explanation
Correct answer: option 1
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
By differentiating both sides
$⇒\frac{dv}{v^2}=-\frac{du}{u^2}$
du : size of object = L
dv : size of image
u : object distance
v : image distance.
$dv=-\frac{v^2}{u^2}du$ $(∵\frac{1}{v}=\frac{1}{f}-\frac{1}{u}⇒v=\frac{fu}{u-f})$
$=-\frac{f^2u^2}{(u-f)^2u^2}du=-\frac{f^2}{(b-f)^2}L$
Negative sign implies that object is lying between u and u + du and the image will lied between v and v - dv.