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A convex spherical mirror with a focal length of magnitude 24.0 cm is placed 20.0 cm to the left of a plane mirror. An object 0.250 cm tall is placed midway between the surface of the plane mirror and the vertex of the spherical mirror. The spherical mirror forms multiple images of the object.
a) Where are the two images of the object formed by the spherical mirror that are closest to the spherical mirror?
b) How tall is each image?

Respuesta :

Answer:

Part a)

First image

[tex]d_i = 7.06 cm[/tex]

second image

[tex]d_i = 13.3 cm[/tex]

Part b)

First image

[tex]h_i = 0.1765 cm[/tex]

Second image

[tex]h_i = 0.11 cm[/tex]

Explanation:

When object is placed between mirror and convex mirror

then for the first image

[tex]d_o = 10 cm[/tex]

[tex]f = 24 cm[/tex]

now from the mirror formula

[tex]\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}[/tex]

here we have

[tex]\frac{1}{d_i} + \frac{1}{-10} = \frac{1}{24}[/tex]

[tex]d_i = 7.06 cm[/tex]

For second position of image we will take object for convex mirror to be the position of image formed in plane mirror

so the distance of that image from convex mirror is

[tex]d_o = 30 cm[/tex]

now by mirror formula

[tex]\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}[/tex]

[tex]\frac{1}{d_i} + \frac{1}{-30} = \frac{1}{24}[/tex]

[tex]d_i = 13.3 cm[/tex]

Part b)

Height of image and height of object is related by the equation

[tex]h_i = \frac{d_i}{d_o} h_o[/tex]

[tex]h_i = \frac{7.06}{10}(0.250)[/tex]

[tex]h_i = 0.1765 cm[/tex]

for height of second image

[tex]h_i = \frac{13.3}{30}(0.250)[/tex]

[tex]h_i = 0.11 cm[/tex]

mickymike92

Based on the mirror formula, the images are at distances of 7.06 cm and 13.3 cm respectively, while the heights of the images are 0.11 cm and 0.05 cm respectively.

What is a convex mirror?

A convex mirror are diverging mirrors which produces virtual, erect and diminished images of object.

Using the mirror formula to determine the distance of the images:

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

where:

  • v is image distance
  • u is object distance
  • f is focal length of mirror

Solving for the distance of the first image:

v = ?

u = -10 cm

f = 24 cm

[tex] \frac{1}{v} + \frac{1}{ - 10} = \frac{1}{24} [/tex]

v = 7.06 cm

Solving for the distance of the second image and taking image in plane mirror as object in spherical mirror:

v = ?

u = -30 cm

f = 24

[tex] \frac{1}{v} + \frac{1}{ - 30} = \frac{1}{24} [/tex]

v = 13.3 cm

Solving for the height of the images:

The heights of the image and object is related as follows:

[tex] \frac{hi}{h} = \frac{v}{u} [/tex]

where:

hi is the image height

h is the object height

Solving for the first image height:

where;

v = 7.06

u = 10 cm

h = 0.25 cm

[tex]hi = \frac{7.06}{10} \times 0.250[/tex]

h = 0.11 cm

For the second image height:

v = 13.3

u = 30 cm

h = 0.11 cm

[tex]hi = \frac{13.36}{30} \times 0.11[/tex]

hi = 0.05 cm

Therefore, the images are at distances of 7.06 cm and 13.3 cm respectively; the heights of the images are 0.11 cm and 0.05 cm respectively.

Learn more about convex mirrors at: https://brainly.com/question/26910366

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