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Solve the following pinhole camera problems: 1) Given u=10cm, Ho=5m, v=2m, calculate Hi and M. 2) Given v=12.67m, Hscreen=16cm, Ho=1.86m, calculate u. 3) Given H_o=3.0m, u=7.50m, calculate image heigh

Physics

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Step-by-step solution

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Here are the solutions to the pinhole camera problems:

1)

  1. Convert all units to meters.

    Object distance (uu) = 10 cm=0.1 m10\, \text{cm} = 0.1\, \text{m}
    Object height (HoH_o) = 5 m5\, \text{m}
    Image distance (vv) = 2 m2\, \text{m}

  2. Calculate the height of the image (HiH_i) using the magnification formula.

    HiHo=vu\frac{H_i}{H_o} = \frac{v}{u} Hi=Ho×vuH_i = H_o \times \frac{v}{u} Hi=5 m×2 m0.1 mH_i = 5\, \text{m} \times \frac{2\, \text{m}}{0.1\, \text{m}} Hi=5 m×20=100 mH_i = 5\, \text{m} \times 20 = \boxed{100\, \text{m}}
  3. Calculate the magnification (MM).

    M=vuM = \frac{v}{u} M=2 m0.1 m=20M = \frac{2\, \text{m}}{0.1\, \text{m}} = \boxed{20}

    2)

  4. Identify the given values and convert units.

    Image distance (vv) = 12.67 m12.67\, \text{m}
    Screen height (HscreenH_{screen}) = 16 cm=0.16 m16\, \text{cm} = 0.16\, \text{m}
    Object height (HoH_o) = 1.86 m1.86\, \text{m}
    For a full photo, the image height (HiH_i) must be equal to the screen height. So, Hi=0.16 mH_i = 0.16\, \text{m}.

  5. Calculate the maximum object distance (uu) using the magnification formula.

    HiHo=vu\frac{H_i}{H_o} = \frac{v}{u} u=v×HoHiu = v \times \frac{H_o}{H_i} u=12.67 m×1.86 m0.16 mu = 12.67\, \text{m} \times \frac{1.86\, \text{m}}{0.16\, \text{m}} u=12.67 m×11.625u = 12.67\, \text{m} \times 11.625 u=147.3 mu = \boxed{147.3\, \text{m}}

    3)

  6. Convert all units to meters.

    Object height (HoH_o) = 3.0 m3.0\, \text{m}
    Object distance (uu) = 7.50 m7.50\, \text{m}
    Image height (HiH_i) = 8.00 cm=0.08 m8.00\, \text{cm} = 0.08\, \text{m}

  7. Calculate the distance of the film from the pinhole (image distance, vv) using the magnification formula.

    HiHo=vu\frac{H_i}{H_o} = \frac{v}{u} v=u×HiHov = u \times \frac{H_i}{H_o} v=7.50 m×0.08 m3.0 mv = 7.50\, \text{m} \times \frac{0.08\, \text{m}}{3.0\, \text{m}} v=7.50 m×0.02666...v = 7.50\, \text{m} \times 0.02666... v=0.20 mv = \boxed{0.20\, \text{m}}

    4)

  8. Convert all units to meters.

    Screen side = 160 mm=0.16 m160\, \text{mm} = 0.16\, \text{m}
    Image distance (vv) = 150 mm=0.15 m150\, \text{mm} = 0.15\, \text{m}
    Object distance (uu) = 11 m11\, \text{m}
    The image occupies three quarters of the height of the screen, so image height (HiH_i) = 34×0.16 m=0.12 m\frac{3}{4} \times 0.16\, \text{m} = 0.12\, \text{m}.

  9. Calculate the height of the flag staff (object height, HoH_o) using the magnification formula.

    HiHo=vu\frac{H_i}{H_o} = \frac{v}{u} Ho=Hi×uvH_o = H_i \times \frac{u}{v} Ho=0.12 m×11 m0.15 mH_o = 0.12\, \text{m} \times \frac{11\, \text{m}}{0.15\, \text{m}} Ho=0.12 m×73.333...H_o = 0.12\, \text{m} \times 73.333... Ho=8.8 mH_o = \boxed{8.8\, \text{m}}

    5)

  10. Identify the given values.

    Length of pinhole camera (image distance, vv) = 0.1 m0.1\, \text{m}
    Object height (HoH_o) = 1 m1\, \text{m}
    Object distance (uu) = 2 m2\, \text{m}

  11. Calculate the height of the image (HiH_i) using the magnification formula.

    HiHo=vu\frac{H_i}{H_o} = \frac{v}{u} Hi=Ho×vuH_i = H_o \times \frac{v}{u} Hi=1 m×0.1 m2 mH_i = 1\, \text{m} \times \frac{0.1\, \text{m}}{2\, \text{m}} Hi=1 m×0.05H_i = 1\, \text{m} \times 0.05 Hi=0.05 mH_i = \boxed{0.05\, \text{m}}
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