An object 1cm tall stands vertically on principal axis of a converging lens, of focal length, f = 1cm, and at a distance of 1.7cm from the lens. Find by graphical construction, the position, size, magnification and nature of the image.

Nguvu ya lenzi ($P$) inatolewa na fomula $P = \frac{1}{f}$, ambapo $f$ lazima iwe katika mita.

Kwanza, badilisha urefu wa focal kutoka sentimita kwenda mita:

$$f = 15 \text{ cm} = 0.15 \text{ m}$$

Sasa, hesabu nguvu ya lenzi:

$$P = \frac{1}{0.15 \text{ m}}$$ $$P = \frac{100}{15} \text{ D}$$ $$P = \frac{20}{3} \text{ D}$$ $$P \approx 6.67 \text{ D}$$

Nguvu ya lenzi ni $\boxed{\textbf{6.67 D}}$

Fomula ya lenzi ni:

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

Tuna $f = 15 \text{ cm}$ na $u = 20 \text{ cm}$. Badilisha maadili haya kwenye fomula:

$$\frac{1}{15} = \frac{1}{v} - \frac{1}{20}$$

Tenga $\frac{1}{v}$:

$$\frac{1}{v} = \frac{1}{15} + \frac{1}{20}$$

Tafuta kigawe cha kawaida cha 15 na 20, ambacho ni 60:

$$\frac{1}{v} = \frac{4}{60} + \frac{3}{60}$$ $$\frac{1}{v} = \frac{7}{60}$$ $$v = \frac{60}{7} \text{ cm}$$ $$v \approx 8.57 \text{ cm}$$

Ukuzaji ($M$) unatolewa na fomula:

$$M = -\frac{v}{u}$$

Tuna $v = \frac{60}{7} \text{ cm}$ na $u = 20 \text{ cm}$. Badilisha maadili haya kwenye fomula:

$$M = -\frac{\frac{60}{7}}{20}$$ $$M = -\frac{60}{7 \times 20}$$ $$M = -\frac{3}{7}$$ $$M \approx -0.43$$

Ukuzaji wa lenzi ni $\boxed{\textbf{-0.43}}$