An object of height 4.25mm is placed at a distance of 10cm from a convex lens of power +5D. Find:
- Focal length of the lens, and
- Size of the image.
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h = 4.25mm = 42.5cm
u = -10cm
P = +5D
$=\frac{1}{20}+\frac{1}{-10}$
$=\frac{-1}{20}$
$\text{v}=-20\text{cm}$
$\text{m}=\frac{\text{h}'}{\text{h}}=\frac{\text{v}}{\text{u}}$
$\frac{\text{h}'}{42.5}=\frac{-20}{-10}$
$\text{h}'=85\text{cm}=8.5\text{mm}$
u = -10cm
P = +5D
- $\text{P}=\frac{1}{\text{f}}$
- $\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}$
$=\frac{1}{20}+\frac{1}{-10}$
$=\frac{-1}{20}$
$\text{v}=-20\text{cm}$
$\text{m}=\frac{\text{h}'}{\text{h}}=\frac{\text{v}}{\text{u}}$
$\frac{\text{h}'}{42.5}=\frac{-20}{-10}$
$\text{h}'=85\text{cm}=8.5\text{mm}$
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