An object is placed 20cm from (a) a converging lens, and (b) a diverging lens, of focal length 15cm. Calculate the image position and magnification in each case.
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u = -20cm
$\frac{1}{\text{v}}-\frac{1}{-20}=\frac{1}{15}$
$\frac{1}{\text{v}}=\frac{1}{15}-\frac{1}{20}$
$\frac{1}{\text{v}}=\frac{1}{60}$
$\text{v}=60\text{cm}$
$\text{m}=\frac{\text{v}}{\text{u}}=\frac{60}{-20}=-3$
$\frac{1}{\text{v}}-\frac{1}{-20}=\frac{1}{-15}$
$\frac{1}{\text{v}}=-\frac{1}{15}-\frac{1}{20}$
$\frac{1}{\text{v}}=-\frac{1}{15}-\frac{1}{20}$
$\frac{1}{\text{v}}=\frac{-7}{60}$
$\text{v}=-8.57\text{cm}$
$\text{m}=\frac{-\text{v}}{\text{u}}=\frac{-8.57}{-20}=+0.42$
- f = 15cm (for converging lens)
$\frac{1}{\text{v}}-\frac{1}{-20}=\frac{1}{15}$
$\frac{1}{\text{v}}=\frac{1}{15}-\frac{1}{20}$
$\frac{1}{\text{v}}=\frac{1}{60}$
$\text{v}=60\text{cm}$
$\text{m}=\frac{\text{v}}{\text{u}}=\frac{60}{-20}=-3$
- f = -15cm(for diverging lens)
$\frac{1}{\text{v}}-\frac{1}{-20}=\frac{1}{-15}$
$\frac{1}{\text{v}}=-\frac{1}{15}-\frac{1}{20}$
$\frac{1}{\text{v}}=-\frac{1}{15}-\frac{1}{20}$
$\frac{1}{\text{v}}=\frac{-7}{60}$
$\text{v}=-8.57\text{cm}$
$\text{m}=\frac{-\text{v}}{\text{u}}=\frac{-8.57}{-20}=+0.42$
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