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Two students in a physics laboratory each have a concave mirror with the same radius of curvature, 55 cm. Each student places an object in front of a mirror. The image in both mirrors is two times the size of the object. However, when the students compare notes, they find that the object distances are not the same. What is the distance of the closer object?

Respuesta :

Answer:

13.75 cm

Explanation:

In a concave mirror , magnified image is formed in two cases

1) When object is placed in between f and 2f . In this case, real image is formed.

2) When object is placed within the focus. In this case virtual image is formed.

Since distance of closer object is asked therefore we shall consider only the second case.

Here v / u = 2

v = 2u

Applying mirror formula

[tex]\frac{1}{v} +\frac{1}{u} =\frac{1}{f} \\[/tex]

f =   -55 / 2 cm.

-[tex][tex]\frac{1}{2u} +\frac{1}{u} = -\frac{2}{55} [/tex][/tex]

u = - 13.75 cm.

onyebuchinnaji

The distance of the closer object is -41.25 cm.

Magnification of each mirror

The magnification of each mirror is calculated as follows;

M = v/u

2 = v/u

v = 2u

where;

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

Distance of the closer object

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

where;

  • f is the focal length = -55 cm/2 = -27.5 cm

[tex]\frac{-1}{27.5} = \frac{1}{2u} + \frac{1}{u} \\\\\frac{-1}{27.5} = \frac{1 + 2}{2u} \\\\\frac{-1}{27.5} = \frac{3}{2u} \\\\2u = -82.5\\\\u = -41.25 \ cm[/tex]

Thus, the distance of the closer object is -41.25 cm.

Learn more about lens equation here: https://brainly.com/question/2860417

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