Pembahasan Uji Prestasi Mandiri 12.3 Sagufindo Kinarya

Soal Nomor 1

Sebuah mikroskop mempunyai lensa objektif dengan kekuatan 100 D dan lensa okuler 25 D. Preparat diletakkan 1,1 cm di bawah lensa objektif. Tentukanlah panjang dan perbesaran mikroskop untuk pengamatan :
a. Tanpa akomodasi
b. Akomodasi maksimum
c. Akomodasi pada jarak 50 cm

Diketahui :
$P_{ob}$ = 100 D
$P_{ok}$ = 25 D
$s_{ob}$ = 1,1 cm = 11 mm

Ditanya :
a. d dan $\gamma$ saat tanpa akomodasi
b. d dan $\gamma$ saat akomodasi
c. d dan $\gamma$ saat akomodasi pada jarak 50 cm

Jawab :
Besar jarak fokus lensa obyektif ($f_{ob}$)
$$\begin{aligned} f_{ob} &= \frac{100}{P_{ob}} \nonumber\\ &= \frac{100}{100} \nonumber\\ &= 1 \ cm \ = \ 10 \ mm \nonumber \end{aligned}$$

Besar jarak fokus lensa okuler ($f_{ok}$)
$$\begin{aligned} f_{ok} &= \frac{100}{P_{ok}} \nonumber\\ &= \frac{100}{25} \nonumber\\ &= 4 \ cm \nonumber \end{aligned}$$

Mencari besar bayangan dari lensa obyektif ($s'_ob$)
$$\begin{aligned} \frac{1}{f_{ob}} &= \frac{1}{s_{ob}} + \frac{1}{s'_{ob}} \nonumber\\ \frac{1}{s'_{ob}} &= \frac{1}{f_{ob}} - \frac{1}{s_{ob}} \nonumber\\ &= \frac{1}{10} - \frac{1}{11} \nonumber\\ &= \frac{11-10}{110} \nonumber\\ &= \frac{1}{110} \nonumber\\ s'_{ob} &= 110 \ mm \ = \ 11 \ cm \nonumber \end{aligned}$$

a. Tanpa akomodasi
$$\begin{aligned} d &= s'_{ob}+f_{ok} \nonumber\\ &= 11 + 4 \nonumber\\ &= 15 \ cm \nonumber\\ \\ \nonumber\\ \gamma &= \left (\frac{s'_{ob}}{s_{ob}}\right) \times \left(\frac{Sn}{f_{ok}}\right) \nonumber\\ &= \left (\frac{11 \ cm}{1,1 \ cm}\right) \times \left(\frac{25 \ cm}{4 \ cm}\right) \nonumber\\ &= 10 \times 6,25 \nonumber\\ &= 62,5 \ kali \nonumber\\ \end{aligned}$$

b. Akomodasi maksimum
    Mencari jarak benda lensa okuler ($s_{ok}$)
$$\begin{aligned} \frac{1}{f_{ok}} &= \frac{1}{s_{ok}} + \frac{1}{s'_{ok}} \nonumber\\ \frac{1}{s_{ok}} &= \frac{1}{f_{ok}} - \frac{1}{s'_{ok}} \nonumber\\ \frac{1}{s_{ok}} &= \frac{1}{f_{ok}} - \frac{1}{-Sn} \nonumber\\ &= \frac{1}{4} +\frac{1}{25} \nonumber\\ &= \frac{25+4}{100} \nonumber\\ &= \frac{29}{100} \nonumber\\ s_{ok} &= \frac{100}{29} = 3,45 \ cm \nonumber\\ \\ \nonumber\\ d &= s'_{ob}+s_{ok} \nonumber\\ &= 11 + 3,45 \nonumber\\ &= 14,45 \ cm \nonumber\\ \\ \nonumber\\ \gamma &= \left (\frac{s'_{ob}}{s_{ob}}\right) \times \left(\frac{Sn}{f_{ok}+1}\right) \nonumber\\ &= \left (\frac{11 \ cm}{1,1 \ cm}\right) \times \left(\frac{25 \ cm}{4 \ cm}+1\right) \nonumber\\ &= 10 \times 7,25 \nonumber\\ &= 72,5 \ kali \nonumber\\ \end{aligned}$$

c. Akomodasi pada jarak 50 cm ($s'_{ok}$ = -50 cm)
    Mencari jarak benda lensa okuler ($s_{ok}$)
$$\begin{aligned} \frac{1}{f_{ok}} &= \frac{1}{s_{ok}} + \frac{1}{s'_{ok}} \nonumber\\ \frac{1}{s_{ok}} &= \frac{1}{f_{ok}} - \frac{1}{s'_{ok}} \nonumber\\ &= \frac{1}{4} -\frac{1}{-50} \nonumber\\ &= \frac{1}{4} +\frac{1}{50} \nonumber\\ &= \frac{25+2}{100} \nonumber\\ &= \frac{27}{100} \nonumber\\ s_{ok} &= \frac{100}{27} = 3,7 \ cm \nonumber\\ \\ \nonumber\\ d &= s'_{ob}+s_{ok} \nonumber\\ &= 11 + 3,7 \nonumber\\ &= 14,7 \ cm \nonumber\\ \\ \nonumber\\ \gamma &= \left (\frac{s'_{ob}}{s_{ob}}\right) \times \left(\frac{Sn}{s_{ok}}\right) \nonumber\\ &= \left (\frac{11 \ cm}{1,1 \ cm}\right) \times \left(\frac{25 \ cm}{3,7 \ cm}\right) \nonumber\\ &= 10 \times 6,76 \nonumber\\ &= 67,6 \ kali \nonumber\\ \end{aligned}$$

Soal Nomor 2

Sebuah mikroskop memiliki jarak fokus okuler 2,5 cm dan jarak fokus lensa obyektif 0,9 cm. Mikroskop ini digunakan oleh orang bermata normal (Sn = 25 cm) tanpa akomodasi. Berapa perbesaran mikroskop jika benda diletakkan 1,0 cm di depan lensa obyektif?

Diketahui :
$f_{ok}$ = 2,5 cm
$f_{ob}$ = 0,9 cm = 9 mm
Sn = 25 cm
tanpa akomodasi
$s_{ob}$ = 1,0 cm = 10 mm

Ditanya :
$\gamma$ = ...?

Jawab :
Mencari jarak bayangan pada lensa obyektif ($s'_{ob}$)
$$\begin{aligned} \frac{1}{f_{ob}} &= \frac{1}{s_{ob}} + \frac{1}{s'_{ob}} \nonumber\\ \frac{1}{s'_{ob}} &= \frac{1}{f_{ob}} - \frac{1}{s_{ob}} \nonumber\\ &= \frac{1}{9} - \frac{1}{10} \nonumber\\ &= \frac{10-9}{90} \nonumber\\ &= \frac{1}{90} \nonumber\\ s'_{ob} &= 90 \ mm \ = \ 9 \ cm \nonumber\\ \end{aligned}$$

Perbesaran mikroskop tanpa akomodasi
$$\begin{aligned} \gamma &= \left (\frac{s'_{ob}}{s_{ob}}\right) \times \left(\frac{Sn}{f_{ok}}\right) \nonumber\\ &= \left (\frac{9 \ cm}{1 \ cm}\right) \times \left(\frac{25 \ cm}{2,5 \ cm}\right) \nonumber\\ &= 9 \times 10 \nonumber\\ &= 90 \ kali \nonumber\\ \end{aligned}$$

Baca Juga Pembahasan Soal Alat-alat Optik:

Soal Latihan 12.1
Soal Latihan 12.2
Uji Prestasi Mandiri 12.1
Soal Latihan 12.3
Uji Prestasi Mandiri 12.2
Soal Latihan 12.4
Uji Prestasi Mandiri 12.3
Soal Latihan 12.5
Uji Prestasi Mandiri 12.4
Soal Latihan 12.6

Soal Nomor 3

Sebuah mikroskop mempunyai lensa objektif ($P_{ob}$ = 12 D) dan lensa okuler ($P_{ok}$ = 30 D yang terpisah sejauh 25 cm. Hitunglah perbesaran mikroskop untuk :
a. Mata berakomodasi maksimum
b. mata tidak berakomodasi

Diketahui :
$P_{ob}$ = 12 D
$P_{ok}$ = 30 D
d = 25 cm

Ditanya :
a. $\gamma_{a}$ = ...?
b. $\gamma_{ta}$ = ...?

Jawab :
Besar jarak fokus lensa obyektif ($f_{ob}$)
$$\begin{aligned} f_{ob} &= \frac{100}{P_{ob}} \nonumber\\ &= \frac{100}{12} \nonumber\\ &= \frac{25}{3} \ cm  \nonumber\\ \end{aligned}$$

Besar jarak fokus lensa okuler ($f_{ok}$)
$$\begin{aligned} f_{ok} &= \frac{100}{P_{ok}} \nonumber\\ &= \frac{100}{30} \nonumber\\ &= \frac {10}{3} \ cm \ = \ 3,33 \ cm \nonumber\\ \end{aligned}$$

a. Perbesaran saat mata berakomodasi maksimum
$$\begin{aligned} d &= s'_{ob}+s_{ok} \nonumber\\ d &= s'_{ob}+(-PP) \nonumber\\ 25 &= s'_{ob} - 25 \nonumber\\ s'_{ob} &= 50 \ cm \nonumber\\ \\ \nonumber\\ \frac{1}{f_{ob}} &= \frac{1}{s_{ob}} + \frac{1}{s'_{ob}} \nonumber\\ \frac{1}{s_{ob}} &= \frac{1}{f_{ob}} - \frac{1}{s'_{ob}} \nonumber\\ &= \frac{1}{\frac{25}{3}} -\frac{1}{50} \nonumber\\ &= \frac{3}{25} - \frac{1}{50} \nonumber\\ &= \frac{(3 \times 4)-2}{100} \nonumber\\ &= \frac{12-2}{100} \nonumber\\ &= \frac{10}{100} \nonumber\\ s_{ob} &= \frac{100}{10} = 10 \ cm \nonumber\\ \\ \nonumber\\ \gamma &= \left (\frac{s'_{ob}}{s_{ob}}\right) \times \left(\frac{Sn}{f_{ok}+1}\right) \nonumber\\ &= \left (\frac{50 \ cm}{10 \ cm}\right) \times \left(\frac{25 \ cm}{\frac{10}{3} \ cm}+1\right) \nonumber\\ &= \left (\frac{50 \ cm}{10 \ cm}\right) \times \left(25 \times \frac{3}{10} +1\right) \nonumber\\ &= 2 \times 8,5 \nonumber\\ &= 17 \ kali \nonumber\\ \end{aligned}$$
 
b. Perbesaran saat mata tidak berakomodasi
$$\begin{aligned} d &= s'_{ob}+f_{ok} \nonumber\\ 25&= s'_{ob} + \frac{10}{3} \nonumber\\ s'_{ob} &= 25 - \frac{10}{3} \nonumber\\ &= \frac{65}{3} \ cm \nonumber\\ &= 21,67 \ cm \nonumber\\ \\ \nonumber\\ \frac{1}{f_{ob}} &= \frac{1}{s_{ob}} + \frac{1}{s'_{ob}} \nonumber\\ \frac{1}{s_{ob}} &= \frac{1}{f_{ob}} - \frac{1}{s'_{ob}} \nonumber\\ &= \frac{1}{\frac{25}{3}} -\frac{1}{\frac{65}{3}} \nonumber\\ &= \frac{3}{25} - \frac{3}{65} \nonumber\\ &= \frac{24}{325} \nonumber\\ s_{ob} &= \frac{325}{24} = 13,54 \ cm \nonumber\\ \\ \nonumber\\ \gamma &= \left (\frac{s'_{ob}}{s_{ob}}\right) \times \left(\frac{Sn}{f_{ok}}\right) \nonumber\\ &= \left (\frac{21,67 \ cm}{13,54 \ cm}\right) \times \left(\frac{25 \ cm}{3,33 \ cm}\right) \nonumber\\ &= 1,6 \times 7,5 \nonumber\\ &= 12 \ kali \nonumber\\ \end{aligned}$$