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A lightbulb is 3m from a wall. What are the focal length and the position (measured from the bulb) of a lens that will form an image on the wall that is twice the size of the lightbulb?

Respuesta :

onyebuchinnaji

The focal length of the lens is 2m and the position of the image is 6 m.

Magnification of the image

The size of the image formed depends on the position of the object from the lens.

[tex]M = \frac{v}{u}[/tex]

where;

  • M is the magnification
  • v is the image distance
  • u is the object distance

Image distance

The image distance is calculated as follows;

[tex]2 = \frac{v}{3} \\\\v = 2(3) \\\\v = 6 \ m[/tex]

Focal length of the lens

The focal length of the lens is calculated as follows;

[tex]\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\\\\\frac{1}{f} = \frac{1}{6} + \frac{1}{3} \\\\\frac{1}{f} = \frac{3}{6} \\\\f = 2m[/tex]

Learn more about image formed by converging lens here: https://brainly.com/question/22394546

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