Tolong ni tolong bgt yakSuatu lensa di udara mempunyai jarak fokus 20 cm. Indeks bias kaca lensanya .... Question from @Nanaaa25

August 2018 1 47 Report

Tolong ni tolong bgt yak

Suatu lensa di udara mempunyai jarak fokus 20 cm. Indeks bias kaca lensanya n = 1,5. berapa dioptri kekuatan lensanya, jika lensa berada dalam bensin? (diketahui indeks bias bensin = 1,2)

please banget kak, ksih jwbn yg bnr ya

Aryasays $\frac{1}{f} =( \frac{nl}{nm} -1)( \frac{1}{R_1} + \frac{1}{R_2} )$
$\frac{1}{20} =( \frac{ \frac{3}{2} }{1} -1)( \frac{1}{R_1} + \frac{1}{R_2} )$
$\frac{1}{20} =( \frac{3}{2} -1)( \frac{1}{R_1} + \frac{1}{R_2} )$
$\frac{1}{20} =( \frac{1}{2})( \frac{1}{R_1} + \frac{1}{R_2} )$
$\frac{1}{10} =\frac{1}{R_1} + \frac{1}{R_2}$ ... (1)

saat dimasukkan ke bensin
$\frac{1}{f} =( \frac{ \frac{3}{2} }{ \frac{6}{5} } -1)( \frac{1}{R_1} + \frac{1}{R_2} )$
$\frac{1}{f} =( \frac{15}{12} -1) \frac{1}{10}$
$\frac{1}{f} =( \frac{3}{12})( \frac{1}{10} )$
$\frac{1}{f} = \frac{3}{120}$
$\frac{1}{f} = \frac{1}{40}$
$f=40$ cm

$P= \frac{100}{f}$
$P= \frac{100}{40}$
$P=2,5$

CMIIW :3

Nanaaa25 3/2 dpt drimna kak?

Aryasays infeks bias lensa 1,5 indeks bias udara 1 jadi 1,5/1 =1,5