An object of height 4cm is placed at a distance of 15cm in front of a concave lens of power, -10 dioptres. Find the size of the image.
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h = 4cm u = -15cm P = -10D$\text{P}=\frac{1}{\text{f}}$
$\text{f}=\frac{1}{\text{P}}$
$=\frac{1}{-10}$
$=-0.1\text{m}$
$=-10\text{cm}$
Using Lens formula,$\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}$
$\frac{1}{\text{v}}=\frac{1}{\text{f}}+\frac{1}{\text{u}}$
$=\frac{1}{-10}+\frac{1}{-15}$
$=\frac{-5}{30}$
$\text{v}=-6\text{cm}$
Magnification, $\text{m}=\frac{\text{h}'}{\text{h}}=\frac{\text{v}}{\text{u}}$$\frac{\text{h}'}{4}=\frac{-6}{-15}$
$\text{h}'=1.6\text{cm}$
$\text{f}=\frac{1}{\text{P}}$
$=\frac{1}{-10}$
$=-0.1\text{m}$
$=-10\text{cm}$
Using Lens formula,$\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}$
$\frac{1}{\text{v}}=\frac{1}{\text{f}}+\frac{1}{\text{u}}$
$=\frac{1}{-10}+\frac{1}{-15}$
$=\frac{-5}{30}$
$\text{v}=-6\text{cm}$
Magnification, $\text{m}=\frac{\text{h}'}{\text{h}}=\frac{\text{v}}{\text{u}}$$\frac{\text{h}'}{4}=\frac{-6}{-15}$
$\text{h}'=1.6\text{cm}$
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