Question

An infinitely long rod lies along the axis of a concave mirror of focal length l. The near end of the rod is at distance u $>$ f from the mirror. Its image will have a length?

• A. $\frac{uf}{u-f}$
• B. $\frac{uf}{u+f}$
• C. $\frac{f^1}{u-f}$
• D. $\frac{f^2}{u-f}$

Answer 1

Answer: (D)

$\frac{f^2}{u-f}$

The mirror formula is given by the equation

$\frac{1}{f}$=$\frac{1}{v}$+$\frac{1}{u}$---------(1)

where f=focal length, v=image distance and u =object distance

As the condition is for mirror its object distance and focal length both lies on left side of mirror thus we consider it as negative where as its image forms behind the mirror we consider it positive.

Hence equation 1 can be reduced as,

$\frac{1}{v}=\frac{1}{u}-\frac{1}{f}$

$v=\frac{uf}{f-u}$-------(2)

Now length of the image $L=|v|-|f|$

$\frac{uf}{f-u}-f=\frac{uf-f^2+uf}{f-u}=\frac{{-f}^2}{f-u}$

$L=\frac{f^2}{u-f}$