Question

4cm high object is placed at a distance of 12cm from a converging lens of focal length of
8cm. Determine the position, size and type of the image.​

Answer 1

Required Answer :-

• Height of object = 4cm

• distance from mirror (u) = 12cm

• Focal length = 8cm

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To Find :-

• The position size and type of image formed.

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

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

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

Answer 2

Answer:-

Object distance (u) = -12 cm

Focal length (f) = 8 cm

Image distance (v) = ?

$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$

$\frac{1}{v } = \frac{1}{f} + \frac{1}{u}$

$\frac{1}{v} = \frac{1}{8} + \frac{1}{ - 12} = \frac{1}{8} - \frac{1}{ 12}$

$= \frac{12 - 8}{8 \times 12}$

$= \frac{4}{96} = \frac{1}{24}$

Therefore the image distance (v) = 24 cm (real image)

Object height (h) = 4 cm

Image height (h') = ?

We know that,

$m = \frac{h'}{h} = \frac{v}{u}$

$\frac{h'}{4} = \frac{24}{ - 12}$

$\frac{h'}{4} = - 2$

$h' = - 2 \times 4$

$h' = - 8 \: cm$

Therefore, image height (h') = -8 cm

Solution :-

Position of the image = 24 cm

Size of the image = -8 cm

Nature of the image = Real, Inverted