The image of a candle flame formed by a lens is obtained on a screen placed on the other side of the lens. If the image is three times the size of the flame and the distance between lens and image is 80cm, at what distance should the candle be placed from the lens? What is the nature of the image at a distance of 80cm and the lens?
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The image is real as only the real image can be taken on the screen.Here, image distance, v = +80cm
Magnification, m = -3
Object distance, u = ?
Now, magnification, $\text{m}=\frac{\text{v}}{\text{u}}\Rightarrow\ -3=\frac{80}{\text{u}}\ \text{or u }=\frac{-80}{3}\text{cm}$
Nature of image will be real, magnified and inverted. The image will be formed beyond 2F.
By using lens formula,
$\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\Rightarrow\ \frac{1}{\text{f}}=\frac{1}{80}-\frac{3}{-80}=\frac{1}{20}\text{ or f }=20\text{cm}$
Focal length is positive so, the lens is convex.
Magnification, m = -3
Object distance, u = ?
Now, magnification, $\text{m}=\frac{\text{v}}{\text{u}}\Rightarrow\ -3=\frac{80}{\text{u}}\ \text{or u }=\frac{-80}{3}\text{cm}$
Nature of image will be real, magnified and inverted. The image will be formed beyond 2F.
By using lens formula,
$\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\Rightarrow\ \frac{1}{\text{f}}=\frac{1}{80}-\frac{3}{-80}=\frac{1}{20}\text{ or f }=20\text{cm}$
Focal length is positive so, the lens is convex.
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