An object is placed at a distance of 100cm from each of the above lenses A and Calculate:
- Image distance, and
- Magnification, in each of the two cases.
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Case 1: For lens A$\text{f}_\text{A}=+50\text{cm}$
$\text{u}_\text{A}=-100\text{cm}$
$\frac{1}{\text{f}_\text{A}}=\frac{1}{\text{v}_\text{A}}-\frac{1}{\text{u}_\text{A}}$
$\frac{1}{\text{v}_\text{A}}=\frac{1}{\text{f}_\text{A}}+\frac{1}{\text{u}_\text{A}}$
$=\frac{1}{50}+\frac{1}{-100}$
$=\frac{1}{100}$
Image distance, $v_A$ = 100cm Magnification, $\text{m}_\text{A}=\frac{\text{v}_\text{A}}{\text{u}_\text{A}}=\frac{100}{-100}=-1$ Case 1: For lens B$\text{f}_\text{B}=-25\text{cm}$
$\text{u}_\text{B}=-100\text{cm}$
$\frac{1}{\text{f}_\text{B}}=\frac{1}{\text{v}_\text{B}}-\frac{1}{\text{u}_\text{B}}$
$\frac{1}{\text{v}_\text{B}}=\frac{1}{\text{f}_\text{B}}+\frac{1}{\text{u}_\text{B}}$
$=\frac{1}{-25}+\frac{1}{-100}$
$=\frac{-5}{100}$
Image distance, $v_B$ = -20cm Magnification, $\text{m}_\text{B}=\frac{\text{v}_\text{B}}{\text{u}_\text{B}}=\frac{-20}{-100}=+0.2$
$\text{u}_\text{A}=-100\text{cm}$
$\frac{1}{\text{f}_\text{A}}=\frac{1}{\text{v}_\text{A}}-\frac{1}{\text{u}_\text{A}}$
$\frac{1}{\text{v}_\text{A}}=\frac{1}{\text{f}_\text{A}}+\frac{1}{\text{u}_\text{A}}$
$=\frac{1}{50}+\frac{1}{-100}$
$=\frac{1}{100}$
Image distance, $v_A$ = 100cm Magnification, $\text{m}_\text{A}=\frac{\text{v}_\text{A}}{\text{u}_\text{A}}=\frac{100}{-100}=-1$ Case 1: For lens B$\text{f}_\text{B}=-25\text{cm}$
$\text{u}_\text{B}=-100\text{cm}$
$\frac{1}{\text{f}_\text{B}}=\frac{1}{\text{v}_\text{B}}-\frac{1}{\text{u}_\text{B}}$
$\frac{1}{\text{v}_\text{B}}=\frac{1}{\text{f}_\text{B}}+\frac{1}{\text{u}_\text{B}}$
$=\frac{1}{-25}+\frac{1}{-100}$
$=\frac{-5}{100}$
Image distance, $v_B$ = -20cm Magnification, $\text{m}_\text{B}=\frac{\text{v}_\text{B}}{\text{u}_\text{B}}=\frac{-20}{-100}=+0.2$
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