Cermin cekung memiliki panjang fokus 5 cm, sebuah benda diletakkan 2 cm didepan cermin itu. berapa p.... Question from @sdwinandaputri

October 2018 2 13 Report

Cermin cekung memiliki panjang fokus 5 cm, sebuah benda diletakkan 2 cm didepan cermin itu. berapa perbesaran bayangan

izharulhaq78 $f=5cm \\ s_o=2cm \\ \\ M=? \\ \\\frac{1}{s_i}=\frac{1}{f}-\frac{1}{s_o} \\ \frac{1}{s_i}=\frac{1}{5}-\frac{1}{2} \\ \frac{1}{s_i}=\frac{2}{10}-\frac{5}{10} \\ \frac{1}{s_i}=-\frac{3}{10} \\s_i=-\frac{10}{3}cm \\ \\ M=| \frac{s_i}{s_o}| \\ M=| \frac{-\frac{10}{3}}{5}| \\ M=|-\frac{10}{3}\times\frac{1}{5}| \\ M=|-\frac{2}{3}\times\frac{1}{1}| \\ M=|-\frac{2}{3}| \\ M=\frac{2}{3}$

0 votes Thanks 1

sdwinandaputri makasih ka:)

echaresya47 Cara cepat :
Si = So x f 2 x 5 10
-------- = --------- = ---- = - 10/3
So - f 2 - 5 -3

M = Si -10/3
---- = ------
So 5
= -10/3 x 1/5
= -2/3

maka M = 2/3 atau 0,66

semoga membantu(:

0 votes Thanks 0

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Sebatang lilin diletakan 8 cm didepan lensa cembung. bayangan yg dihasilkan yaitu maya, maya tegak yg diperbesar 2 kali. hitung jarak fokus lensa!

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