Physics - Geometrical Optics Question with Solution | TestHub

$u_B=x+f=20\mathrm{cm}=R$, so the image of B is at B itself. For end A, $u_A=\ell +x+f=30\mathrm{cm}$.

$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{1}{10}-\frac{1}{30}=\frac{2}{30}\Rightarrow v=15\mathrm{cm}$. Image length $=x+f-v=20-15=5.0\mathrm{cm}$.

Lateral magnification at A is $\frac{15}{30}=\frac{1}{2}$. With $r_A=\frac{d}{2}=1.0\mathrm{mm}$, $r_A^{\prime }=0.5\mathrm{mm}$, so end A's image is $2r_A^{\prime }=1.0\mathrm{mm}$ thick. Option (A) is true.