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You have a piece of film paper that is 3 in x 5 in. You fix it inside the back of a pinhole camera with a focal length of 5.5 in. You want to use it to take a picture of your team’s mascot – a giant guinea pig that just barely fits in a 4 ft. tall cube. The picture will be taken directly in front, from a stool that places the aperture 2 ft. above the ground. You have to determine how far away the camera must be from your mascot to get a good portrait that fills up the whole film paper, without cutting any part of him off. How far apart should the camera and the mascot be to take the portrait? Show your work.

Respuesta :

Answer:

In order to take a portrait, the distance of the mascot from the camera should be approximately 7.33 feet

Explanation:

The size of the film paper = 3 in. × 5 in.

The focal length of the camera = 5.5 in.

The height and width of the guinea pig = 4 ft.

The height of the aperture above the ground = 2 ft.

Therefore, we have;

Magnification = Height of image/(Height of object)

Withe the 3 in. wide film, we have;

Magnification = 3 in./(4 ft.) = 3 in./(48 in.) = 0.0625

Magnification = Length of camera/(Distance of object from pin hole)  

∴ Length of camera/(Distance of object from pin hole) = 0.0625

Length of camera = Focal length of the camera = 5.5 in.

Therefore;

5.5 in./(Distance of object from pin hole) = 0.0625

Distance of object from pin hole = 5.5/0.0625 = 88 inches = 7.33 ft

Therefore, the camera should be approximately 7.33 ft. from the mascot to take a portrait.

lhmarianateixeira

In this exercise we have to use the magnification knowledge to calculate the distance that the photograph should be taken, thus we have to:

Distance of the mascot from the camera should be approximately 7.33 feet

To calculate the best distance to take the photo, we have that some information must be taken into account such as:

  • Size of the film paper: [tex](3)*(5) in[/tex]
  • Focal length of the camera: [tex]5.5 in[/tex]
  • Height and width of the guinea pig: [tex]4 ft[/tex]
  • Height of the aperture above the ground: [tex]2 ft[/tex]

Therefore, we have that the formula of magnification is:

[tex]Magnification = Height \ of \ image/(Height \ of \ object)[/tex]

With the 3 in wide film, we have;

[tex]Magnification = 3 in/(4 ft) \\= 3 in/(48 in) = 0.0625 in[/tex]

Rewriting the magnification formula as:

[tex]Magnification = Length \ of \ camera/(Distance \ of \ object \ from \ pin \ hole)[/tex]  

Substituting the values ​​already known we have the equation will be matched as:

[tex]Length\ of \ camera/(Distance\ of\ object \ from \ pin \ hole) = 0.0625\\Length \ of \ camera = Focal \ length \ of \ the \ camera = 5.5 in.[/tex]

Therefore;

[tex]5.5 /(Distance \ of \ object\ from \ pin\ hole) = 0.0625 in\\Distance \ of \ object\ from \ pin \ hole = 5.5/0.0625\\ = 88 inches = 7.33 ft[/tex]

See more about distance at brainly.com/question/989117

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