4cm high object is placed at a distance of 12cm from a converging lens

1.	4cm high object is placed at a distance of 12cm from a converging lens of focal length of8cm. Determine the position, size and type of the image.
Answer» Required Answer :- Height of object = 4cm distance from MIRROR (u) = 12cm Focal length = 8cm _____________________________ To Find :- The POSITION size and type of image formed. ___________________________________________ $\frac{1}{f} = \frac{1}{u} + \frac{1}{v} \\ \\ \frac{1}{8} = \frac{1}{12} + \frac{1}{v} \\ \\ \frac{1}{v} = \frac{1}{ - 24} - \frac{1}{ - 12} \\ \\ v = 24cm$ $m = - \frac{v}{u} \\ \\ m = - \frac{ - 24}{ - 3} \\ \\ m = 8cm$ Hence Image formed is virtual and ERECT and position is behind the mirror.

4cm high object is placed at a distance of 12cm from a converging lens of focal length of8cm. Determine the position, size and type of the image.

Answer»

Required Answer :-

Height of object = 4cm

distance from MIRROR (u) = 12cm

Focal length = 8cm

_____________________________

To Find :-

The POSITION size and type of image formed.

___________________________________________

$\frac{1}{f} = \frac{1}{u} + \frac{1}{v} \\ \\ \frac{1}{8} = \frac{1}{12} + \frac{1}{v} \\ \\ \frac{1}{v} = \frac{1}{ - 24} - \frac{1}{ - 12} \\ \\ v = 24cm$

$m = - \frac{v}{u} \\ \\ m = - \frac{ - 24}{ - 3} \\ \\ m = 8cm$

Hence Image formed is virtual and ERECT and position is behind the mirror.

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4cm high object is placed at a distance of 12cm from a converging lens of focal length of8cm. Determine the position, size and type of the image.​