An object 2cm tall stands on the principal axis of a converging lens of focal length 8cm. Find the position, nature and size of the image formed if the object is:
- 12cm from the lens.
- 6 cm from the lens.
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$h_1 = 2cm$
f = 8cm
$\frac{1}{\text{v}}-\frac{1}{12}=\frac{1}{8}$
$\frac{1}{\text{v}}=\frac{1}{24}$
$\text{v}=24\text{cm}$
Image is 24cm behind the lens.
$\text{m}=\frac{\text{v}}{\text{u}}=\frac{\text{h}_2}{\text{h}_1}$
$\frac{24}{-12}=\frac{\text{h}_2}{2}$
$\text{h}_2=-4\text{cm}$
Image is 4cm hight, real and inverted.
$\frac{1}{\text{v}}-\frac{1}{-6}=\frac{1}{8}$
$\frac{1}{\text{v}}=-\frac{1}{24}$
$\text{v}=-24\text{cm}$
Image is 24cm in front of the lens.
$\text{m}=\frac{\text{v}}{\text{u}}=\frac{\text{h}_2}{\text{h}_1}$
$\frac{-24}{-6}=\frac{\text{h}_2}{2}$
$\text{h}_2=8\text{cm}$
Image is 8cm high, virtual and erect.
f = 8cm
- u = -12cm
$\frac{1}{\text{v}}-\frac{1}{12}=\frac{1}{8}$
$\frac{1}{\text{v}}=\frac{1}{24}$
$\text{v}=24\text{cm}$
Image is 24cm behind the lens.
$\text{m}=\frac{\text{v}}{\text{u}}=\frac{\text{h}_2}{\text{h}_1}$
$\frac{24}{-12}=\frac{\text{h}_2}{2}$
$\text{h}_2=-4\text{cm}$
Image is 4cm hight, real and inverted.
- u = -6cm
$\frac{1}{\text{v}}-\frac{1}{-6}=\frac{1}{8}$
$\frac{1}{\text{v}}=-\frac{1}{24}$
$\text{v}=-24\text{cm}$
Image is 24cm in front of the lens.
$\text{m}=\frac{\text{v}}{\text{u}}=\frac{\text{h}_2}{\text{h}_1}$
$\frac{-24}{-6}=\frac{\text{h}_2}{2}$
$\text{h}_2=8\text{cm}$
Image is 8cm high, virtual and erect.
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