NTSE Stage 2 SAT 2019 Exam Question Paper

For the end B, image distance of end B will be,
$\mathit{f}=10\mathit{c}\mathit{m}$
${\mathit{u}}_{\mathit{B}}=-18\mathit{c}\mathit{m}$
${\mathit{v}}_{\mathit{B}}=$ image distance of end B
As we know,
$\frac{1}{\mathit{f}}=\frac{1}{{\mathit{v}}_{\mathit{B}}}-\frac{1}{{\mathit{u}}_{\mathit{B}}}$
$\frac{1}{{\mathit{v}}_{\mathit{B}}}=\frac{1}{\mathit{f}}+\frac{1}{{\mathit{u}}_{\mathit{B}}}$
$\frac{1}{{\mathit{v}}_{\mathit{B}}}=\frac{1}{10}-\frac{1}{18}=\frac{8}{180}$
${\mathit{v}}_{\mathit{B}}=\frac{180}{8}\Rightarrow 22.5\mathit{c}\mathit{m}$
Similarly, for the end A, image distance of end A will be,
$\mathit{f}=10\mathit{c}\mathit{m}$
${\mathit{u}}_{\mathit{A}}=-20\mathit{c}\mathit{m}$
${\mathit{v}}_{\mathit{A}}=$ image distance of end A
$\frac{1}{\mathit{f}}=\frac{1}{{\mathit{v}}_{\mathit{A}}}-\frac{1}{{\mathit{u}}_{\mathit{A}}}$
$\frac{1}{{\mathit{v}}_{\mathit{A}}}=\frac{1}{\mathit{f}}+\frac{1}{{\mathit{u}}_{\mathit{A}}}$
$\frac{1}{{\mathit{v}}_{\mathit{A}}}==\frac{1}{10}-\frac{1}{20}=\frac{1}{20}$
${\mathit{v}}_{\mathit{A}}=20\mathit{c}\mathit{m}$
So,length of image $\mathit{A}\prime \mathit{B}\prime ={\mathit{v}}_{\mathit{B}}-{\mathit{v}}_{\mathit{A}}$
$=22.5-20=2.5$cm
So magnification ,$\mathit{m}=\frac{\mathit{A}\prime \mathit{B}\prime }{\mathit{A}\mathit{B}}\Rightarrow \frac{2.5}{2}=1.25$