A convex lens has a focal length of 10cm. Find the location and nature of the image if a point object is placed on the principal axis at a distance of:
  1. 9.8cm
  2. 10.2cm from the lens.
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Given that, f = 10cm
  1. When u = -9.5cm
$\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{F}}\Rightarrow\frac{1}{\text{v}}=\frac{1}{10}-\frac{1}{9.8}=\frac{-0.2}{98}$

$\Rightarrow\text{v}=-490\text{cm}$

So, $\text{m}=\frac{\text{v}}{\text{u}}=\frac{-490}{-9.8}=50\text{cm}$

So, the image is erect and virtual.
  1. When u = -10.2cm
$\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{F}}\Rightarrow\frac{1}{\text{v}}=\frac{1}{10}-\frac{1}{-10.2}=\frac{102}{0.2}$

$\Rightarrow\text{v}=510\text{cm}$

So, $\text{m}=\frac{\text{v}}{\text{u}}=\frac{510}{-9.8}$

The image is real and inverted.
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