Supiyanto Kelas X Evaluasi Bab 5 Nomor 8

Titik jauh penglihatan seseorang 100 cm di depan mata. Orang ini memerlukan kacamata dengan lensa berdaya …

A. +0,3 D D. -1 D
B. +0,5 D E. -3 D
C. +3 D

Penyelesaian

Fokus lensa

\frac{1}{f}=\frac{1}{\infty }-\frac{1}{100}

\frac{1}{f}=-\frac{1}{100}\frac{1}{\text{cm}}

 

Daya lensa

P=\frac{100}{-100}=-1~\text{D}

Jawaban D

Supiyanto. 2006. FISIKA UNTUK SMA KELAS X. Jakarta: Phiβeta.

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