Q16: The image of a candle flame placed at a distance of 30 cm from a mirror is formed on a screen placed in front of the mirror at a distance of 60 cm from its pole. What is the nature of the mirror? Find its focal length. If the height of the flame is 2.4 cm, find the height of its image. State whether the image formed is erect or inverted.

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Solution

Dear Student ,

$u = 30 c m, v = 60 c m, h = 2.4 c m \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \frac{1}{f} = \frac{1}{60} + \frac{1}{30} f = 20 c m \frac{h'}{h} = \frac{v}{u} \frac{h'}{2.4} = - 2 h' = - 4.8 C o n v e x l e n s f = 20 c m h' = - 4.8 c m$
Regards

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Dear Student , u=30 cm , v=60 cm ,h=2.4cm 1f=1v-1u1f=160+130f=20cm h'h=vuh'2.4=-2h'=-4.8Convex lens f=20 cm h'=-4.8cm Regards

Dear Student ,

$u = 30 c m, v = 60 c m, h = 2.4 c m \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \frac{1}{f} = \frac{1}{60} + \frac{1}{30} f = 20 c m \frac{h'}{h} = \frac{v}{u} \frac{h'}{2.4} = - 2 h' = - 4.8 C o n v e x l e n s f = 20 c m h' = - 4.8 c m$
Regards