An object is placed at a distance 24 cm in front of a convex lens of focal length 8 cm.
(i) What is the nature of the image so formed ?
(ii) Calculate the distance of the image from the lens.
(iii) Calculate the magnification of the image.
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Given u = 24 cm (-ve), f = 8 cm (+ve)
(i) Image formed is real, inverted and diminished.
(ii) Let v be the distance of image from optical centre. From the relation,
$$\begin{aligned}\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \\[0.5em]\end{aligned}$$
Substituting the values we get,
$$\begin{aligned}\frac{1}{v} + \frac{1}{24} = \frac{1}{8} \\[0.5em] \frac{1}{v} = \frac{1}{8} - \frac{1}{24} \\[0.5em] \frac{1}{v} = \frac{3 - 1}{24} \\[0.5em] \frac{1}{v} = \frac{2}{24} \\[0.5em] \frac{1}{v} = \frac{1}{12} \\[0.5em] \Rightarrow v = 12 \text{cm}\end{aligned}$$
Hence, the distance of the image from the lens = 12 cm
(iii) From relation, m = $\frac{v}{u}$
Substituting the values we get,
$$\begin{aligned}m = \frac{12}{-24} \\[0.5em] \Rightarrow m = -0.5\end{aligned}$$
-ve sign of magnification tells us that the image is inverted.
Hence, magnification = -0.5