The image of a candle flame placed at a distance of 45 cm from aspherical lens is formed on screen placed at a distance of 90 cm from the lens. Identify the type of lens and calculate its focal length If the height of the flames 2cm, find the height of its image.
CBSE DELHI - SET 1 2012
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Convex lens$\frac{1}{f} = \frac{1}{v}-\frac{1}{u}$
$\frac{1}{f} =\frac{1}{(\text{+ 90)}}-\frac{1}{(\text{- 45)}}$
$ f\text{ =} + 30 \text{ cm}$
Now,$m= \frac{h'}{h} = \frac{v}{u}$
$m= \frac{h'}{2} = \frac{90}{-45}$
$\therefore h' = - \text{ 4} \text{ cm}$
Height of flame image is -4cm.
$\frac{1}{f} =\frac{1}{(\text{+ 90)}}-\frac{1}{(\text{- 45)}}$
$ f\text{ =} + 30 \text{ cm}$
Now,$m= \frac{h'}{h} = \frac{v}{u}$
$m= \frac{h'}{2} = \frac{90}{-45}$
$\therefore h' = - \text{ 4} \text{ cm}$
Height of flame image is -4cm.
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