Question
An object is 2m from a lens which forms an erect image one-fourth (exactly) the size of the object. Determine the focal length of the lens. What type of lens is this?
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Answer
Given, Object distance (u) = -2m Image distance (v) = ? Magnification (m) = 0.25 (one-fourth of the size of the image) Focal length (f) = ?$\text{Magnification(m)}=\frac{\text{v}}{\text{u}}$
$\text{v}=-0.5\text{m}$
Putting these values in lens formula, we get,$\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\Rightarrow\frac{1}{-0.5}-\frac{1}{-2}=\frac{1}{\text{f}}$
$\Rightarrow\frac{1}{-0.5}+\frac{1}{2}=\frac{1}{\text{f}}$
$\Rightarrow\frac{1}{\text{f}}=\frac{1}{2}-\frac{1}{0.5}$
$\Rightarrow\frac{1}{\text{f}}=\frac{1}{2}-\frac{10}{5}$
$\Rightarrow\frac{1}{\text{f}}=\frac{-3}{2}$
$\Rightarrow\text{f}=-0.66\text{m}$
Negative sign of focal length shows that lens is diverging in nature. Hence, it is a concave lens.
$\text{v}=-0.5\text{m}$
Putting these values in lens formula, we get,$\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\Rightarrow\frac{1}{-0.5}-\frac{1}{-2}=\frac{1}{\text{f}}$
$\Rightarrow\frac{1}{-0.5}+\frac{1}{2}=\frac{1}{\text{f}}$
$\Rightarrow\frac{1}{\text{f}}=\frac{1}{2}-\frac{1}{0.5}$
$\Rightarrow\frac{1}{\text{f}}=\frac{1}{2}-\frac{10}{5}$
$\Rightarrow\frac{1}{\text{f}}=\frac{-3}{2}$
$\Rightarrow\text{f}=-0.66\text{m}$
Negative sign of focal length shows that lens is diverging in nature. Hence, it is a concave lens.
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