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Focal length 20. Object 10, 30, 40, 50, 60.

Physics

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Huu unaonekana kuwa mfumo wa lenzi au kioo. Tutatumia fomula ya lenzi/kioo kupata umbali wa picha (vv) kwa kila umbali wa kitu (uu). Focal length (ff) imetolewa kama 20.

Fomula ya lenzi/kioo ni:

1f=1u+1v\frac{1}{f} = \frac{1}{u} + \frac{1}{v}

Ili kupata vv, tunapanga upya fomula:

1v=1f−1u\frac{1}{v} = \frac{1}{f} - \frac{1}{u}

Ambapo f=20f = 20.

  1. Tafuta vv wakati u=10u = 10.

    1v=120−110\frac{1}{v} = \frac{1}{20} - \frac{1}{10} 1v=120−220\frac{1}{v} = \frac{1}{20} - \frac{2}{20} 1v=−120\frac{1}{v} = -\frac{1}{20} v=−20v = -20
  2. Tafuta vv wakati u=30u = 30.

    1v=120−130\frac{1}{v} = \frac{1}{20} - \frac{1}{30} 1v=360−260\frac{1}{v} = \frac{3}{60} - \frac{2}{60} 1v=160\frac{1}{v} = \frac{1}{60} v=60v = 60
  3. Tafuta vv wakati u=40u = 40.

    1v=120−140\frac{1}{v} = \frac{1}{20} - \frac{1}{40} 1v=240−140\frac{1}{v} = \frac{2}{40} - \frac{1}{40} 1v=140\frac{1}{v} = \frac{1}{40} v=40v = 40
  4. Tafuta vv wakati u=50u = 50.

    1v=120−150\frac{1}{v} = \frac{1}{20} - \frac{1}{50} 1v=5100−2100\frac{1}{v} = \frac{5}{100} - \frac{2}{100} 1v=3100\frac{1}{v} = \frac{3}{100} v=1003≈33.33v = \frac{100}{3} \approx 33.33
  5. Tafuta vv wakati u=60u = 60.

    1v=120−160\frac{1}{v} = \frac{1}{20} - \frac{1}{60} 1v=360−160\frac{1}{v} = \frac{3}{60} - \frac{1}{60} 1v=260\frac{1}{v} = \frac{2}{60} 1v=130\frac{1}{v} = \frac{1}{30} v=30v = 30

    Majibu ya umbali wa picha (vv) kwa kila umbali wa kitu (uu) ni:

    • Wakati u=10u = 10, v=−20v = \boxed{-20}
    • Wakati u=30u = 30, v=60v = \boxed{60}
    • Wakati u=40u = 40, v=40v = \boxed{40}
    • Wakati u=50u = 50, v=33.33v = \boxed{33.33}
    • Wakati u=60u = 60, v=30v = \boxed{30}
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