Question

A rod of length $\ell =10cm$ and diameter $d=2.0mm$ is placed along the axis of a concave mirror of focal length $f=10cm$ as shown in the figure. End $B$ at distance $x=10cm$ from the focus $F$ of the mirror. Then

• A. Length of the image is $5.0cm$ and end $A$ is $1.0mm$ thick
• B. Length of the image is $6.0cm$ and end $A$ is $1.0mm$ thick
• C. Length of the image is $5.0cm$ and end $A$ is $2.0mm$ thick
• D. Length of the image is $6.0cm$ and end $A$ is $4.0mm$ thick

Answer 1

Answer: (A)

The image of $B$ is $B$ itself
since $U_B=x+f=20cm=R$ for end $A$,
$U_A=l+x+f=30cm$
Its image distance is $\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$
$=-\frac{1}{10}+\frac{1}{30}=\frac{2}{30}$
$\Rightarrow v=-15cm$
so the image length $=x+f-v=5.0cm$
lateral magnification of $B$ in $1.0$
and that of $A$ is $\frac{15}{30}=\frac{1}{2}=\frac{r_A^1}{r_A}=r_A^1=1.0mm$
so, option (a) is true.