1.20) x1=-1m= -100cm
[tex] \frac{1}{x2} - \frac{1}{x1} = \frac{1}{f} [/tex]
[tex] \frac{1}{x2} + \frac{1}{100} = \frac{1}{50} [/tex]
[tex] \frac{100 + x2}{100 \times x2} = \frac{1}{50} [/tex]
50(100+x2)=100×x2
5000+50x2=100x2
5000=50x2
x2=5000/50
x2=100 cm
Imaginea se afla la 100 cm de lentila.
marirea liniara=x2/x1= 100/-100=-1 => imaginea este reala si egala cu obiectul.
1.21) x1=-40 cm
f=-20 cm (lentila divergenta)
[tex] \frac{1}{ x2 } + \frac{1}{40} = - \frac{1}{20} [/tex]
[tex] \frac{40 + x2}{40 \times x2} = - \frac{1}{20} [/tex]
-20(40+x2)=-1(40x2)
-800-20x2=-40x2 / +20x2
-800=-20x2
x2=40 cm
x2/x1=40/-40=-1 => imagine reala,la 40 cm de lentila
1.22) y1=6m
y2=24mm
y1=600 cm
y2=2.4cm
y2/y1=0.004
x2/x1= 0.004 => x2=0.004x1
[tex] \frac{x1 - x2}{x1x2} = \frac{1}{5} [/tex]
5(x1-0.004x1)=0.004 x1 la patrat
5x1-0.02x1=0.004 x1 la patrat
4.98x1=0.004x1 la patrat /÷x
4.98=0.004 x1
x1=1245 cm=12.45 m
obiectul de afla la 12.45 m de lentila
