Question

An object of height 4.25 mm is placed at a distance of 10 cm from a convex lens of power +5D. Find (i) focal length of the lens, and (ii) size of the image.

Answer 1

Object height, h = 4.25 mm = 0.425 cm (1 cm = 10 mm)
Object distance, u = $-$10 cm
Power, P = +5 D
Focal length, f = ?
Image distance, v = ?
Image height, h' = ?

Power, P = $\frac{1}{f}$

f = $\frac{1}{\mathrm{p}}=\frac{1}{5}=0.2\mathrm{m}=20\mathrm{cm}$

Using the lens formula, we get:

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

$$\begin{gathered} \Rightarrow \frac{1}{v}-\frac{1}{-10}=\frac{1}{20} \\ \Rightarrow \frac{1}{v}+\frac{1}{10}=\frac{1}{20} \\ \Rightarrow \frac{1}{v}=\frac{1}{20}-\frac{1}{10} \\ \Rightarrow \frac{1}{v}=-\frac{1}{20} \\ \therefore v\mathit{}=-20\mathrm{cm} \end{gathered}$$

Now, magnification, m = $\frac{\mathit{h}\mathit{\text{'}}}{\mathit{h}}\mathit{=}\frac{\mathit{v}}{\mathit{u}}$

Substituting the values in the above equation, we get:

$$\begin{gathered} \frac{h\mathit{\text{'}}}{0.425}=\frac{-20}{-10} \\ h\mathit{\text{'}}=2\mathrm{x}0.425=0.85\mathrm{cm}=8.5\mathrm{mm} \end{gathered}$$

Thus, the image is 8.5 mm long; it is also erect and virtual.