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A 2cm high arrow is placed 24 cm in front of a concave mirror that has a focal length with a magnitude of 17cm.

Find:
1. The image distance.
2. The magnification of the mirror in this setup.
3. The image height.

Repeat parts A through C for the case where the arrow is now placed 11 cm from the mirror.

Answer :

Let's solve the problem using the mirror equation and magnification formula.

Definitions


  • Concave mirror: A mirror that curves inward, like the inside of a sphere.

  • Focal length (f): The distance from the mirror to the focal point.

  • Object distance (d_o): The distance from the mirror to the object.

  • Image distance (d_i): The distance from the mirror to the image.

  • Magnification (m): The ratio of the height of the image to the height of the object.


We use the mirror equation to find the image distance:

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

And the magnification formula:

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

where:


  • [tex]h_i[/tex] is the image height,

  • [tex]h_o[/tex] is the object height.


Given:


  • Object height, [tex]h_o[/tex]: 2 cm

  • Object distance, [tex]d_o[/tex]: 24 cm

  • Focal length, [tex]f[/tex]: -17 cm (for concave mirrors)


1. Calculate the image distance [tex]d_i[/tex]

Using the mirror equation:
[tex]\frac{1}{-17} = \frac{1}{24} + \frac{1}{d_i}[/tex]

Rearrange to find [tex]\frac{1}{d_i}[/tex]:
[tex]\frac{1}{d_i} = \frac{1}{-17} - \frac{1}{24} = \frac{-24 - 17}{408}[/tex]
[tex]\frac{1}{d_i} = \frac{-41}{408}[/tex]

So,
[tex]d_i = \frac{408}{-41} \approx -9.95 \text{ cm}[/tex]

2. Calculate the magnification [tex]m[/tex]

Using the magnification formula:
[tex]m = -\frac{d_i}{d_o} = -\frac{-9.95}{24} \approx 0.414[/tex]

3. Calculate the image height [tex]h_i[/tex]

Using the magnification formula:
[tex]h_i = m \times h_o = 0.414 \times 2 \text{ cm} \approx 0.828 \text{ cm}[/tex]

The image is upright and smaller than the object.

Now, repeat the steps for the object placed 11 cm from the mirror.

1. Calculate the image distance [tex]d_i[/tex]

Using:
[tex]\frac{1}{-17} = \frac{1}{11} + \frac{1}{d_i}[/tex]

Rearrange:
[tex]\frac{1}{d_i} = \frac{-1}{17} - \frac{1}{11} = \frac{-11 - 17}{187}[/tex]
[tex]\frac{1}{d_i} = \frac{-28}{187}[/tex]

So,
[tex]d_i = \frac{187}{-28} \approx -6.68 \text{ cm}[/tex]

2. Calculate the magnification [tex]m[/tex]

[tex]m = -\frac{d_i}{d_o} = -\frac{-6.68}{11} \approx 0.607[/tex]

3. Calculate the image height [tex]h_i[/tex]

[tex]h_i = m \times h_o = 0.607 \times 2 \text{ cm} \approx 1.214 \text{ cm}[/tex]

The image is also upright, but less enlarged compared to when it was 24 cm from the mirror.

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