Question

Ex 1.20,1.21,1.22

Dau coroană:)

Answer 1

1.20) x1=-1m= -100cm

$\frac{1}{x2} - \frac{1}{x1} = \frac{1}{f}$

$\frac{1}{x2} + \frac{1}{100} = \frac{1}{50}$

$\frac{100 + x2}{100 \times x2} = \frac{1}{50}$

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)

$\frac{1}{ x2 } + \frac{1}{40} = - \frac{1}{20}$

$\frac{40 + x2}{40 \times x2} = - \frac{1}{20}$

-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

$\frac{x1 - x2}{x1x2} = \frac{1}{5}$

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