An object is placed at a distance of $100 \text{ cm}$ from a converging lens of focal length $40 \text{ cm}$. $(i)$ What is the nature of the image? (ii) What is the position of the image? A convex lens produces an inverted image magnified three times of an object placed at a distance of $15 \text{ cm}$ from it. Calculate the focal length of the lens.

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(N/A) Part $1$: Given $u = -100 \text{ cm}$, $f = +40 \text{ cm}$.
Using lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$:
$\frac{1}{v} - \frac{1}{-100} = \frac{1}{40} \implies \frac{1}{v} = \frac{1}{40} - \frac{1}{100} = \frac{5-2}{200} = \frac{3}{200}$.
$v = \frac{200}{3} \approx 66.67 \text{ cm}$. Since $v$ is positive, the image is real and inverted.
Part $2$: Given $u = -15 \text{ cm}$, magnification $m = -3$ (inverted image).
$m = \frac{v}{u} \implies -3 = \frac{v}{-15} \implies v = 45 \text{ cm}$.
Using $\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{45} - \frac{1}{-15} = \frac{1+3}{45} = \frac{4}{45}$.
$f = \frac{45}{4} = 11.25 \text{ cm}$.

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