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An object of height $$\begin{aligned}& 6 \\ & \text{cm}\end{aligned}$$ is placed at a distance of $$\begin{aligned}& 30 \\ & \text{cm}\end{aligned}$$ from the optical centre of a concave lens of focal length $$\begin{aligned}& 15 \\ & \text{cm}\end{aligned}$$.
Use lens formula to determine :
(a) The distance of image from the optical centre.
(b) The height of the image formed.
Use lens formula to determine :
(a) The distance of image from the optical centre.
(b) The height of the image formed.
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(a) Height of object = $\text{6.0 cm}$
$u = \text{-30 cm}$
$f = \text{-15 cm}$
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$\frac{1}{v} = \frac{1}{f} + \frac{1}{u}$
$\frac{1}{v} = \frac{1}{-15} + \frac{1}{-30}$ (1/2)
(b) $v = \text{-10 cm}$
$m = \frac{v}{u}$
$m = \frac{-10}{-30}$
$m = \frac{1}{3}$
$\text{height of image} = \frac{h'}{\text{height of object}} = \frac{h'}{h}$
$h' = m \times h$
$h' = \frac{1}{3} \times 6$
$h' = \text{2 cm}$ (1/2)
$u = \text{-30 cm}$
$f = \text{-15 cm}$
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$\frac{1}{v} = \frac{1}{f} + \frac{1}{u}$
$\frac{1}{v} = \frac{1}{-15} + \frac{1}{-30}$ (1/2)
(b) $v = \text{-10 cm}$
$m = \frac{v}{u}$
$m = \frac{-10}{-30}$
$m = \frac{1}{3}$
$\text{height of image} = \frac{h'}{\text{height of object}} = \frac{h'}{h}$
$h' = m \times h$
$h' = \frac{1}{3} \times 6$
$h' = \text{2 cm}$ (1/2)