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.

Huu hapa ni utatuzi wa swali la 5:

5. An object 5cm high is placed 20cm in front of a converging lens of focal length 15cm. Find the power of the lens and the magnification of the lens.

Tunapewa: Urefu wa kitu, $h_o = 5 \text{ cm}$ Umbali wa kitu, $u = 20 \text{ cm}$ Urefu wa focal, $f = 15 \text{ cm}$ (kwa lenzi inayokusanya, $f$ ni chanya)

Step 1: Tafuta Nguvu ya Lenzi (Power of the lens) 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: $f = 15 cm = 0.15 \text{ m}$

Sasa, hesabu nguvu ya lenzi: $P = \frac{1}{0.15 m}$ $P = \frac{100}{15} \text{ D}$ $P = \frac{20}{3} \text{ D}$ $P \approx 6.67 \text{ D}$ Nguvu ya lenzi ni $\boxed{6.67 D}$

Step 2: Tafuta Umbali wa Picha ($v$) kwa kutumia Fomula ya Lenzi 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: $\frac{1}{v} = \frac{4}{60} + \frac{3}{60}$ $\frac{1}{v} = \frac{7}{60}$ $v = \frac{60}{7} \text{ cm}$ $v \approx 8.57 \text{ cm}$

Step 3: Tafuta Ukuzaji wa Lenzi (Magnification of the lens) 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 $\boxed{-0.43}$

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