Question

The near end of the rod which lies along the axis of a concave mirror of focal length f is at a distance $u>f$ from the mirror. The length of the image is $\frac{f^2}{(u-f)}$. The far end of the rod will lie at :

• A. f
• B. 2 f
• C. 4 f
• D. Infinity

Answer 1

The correct option is D Infinity


$V_1=\frac{(-u)(-f)}{(-u)-(-f)}$
$V_1=\frac{uf}{f-u}$
$V_1=\frac{uf}{f-u}$
$|V_1-V_2|=\frac{f^2}{u-f}$
$\frac{uf}{f-u}-V_2=\frac{f^2}{f-u}$
$V_2=\frac{uf-f^2}{f-u}$
$V_2=\frac{f(u-f)}{f-u}$
$V_2=-f$
$u_2=\infty$