COLLEGE PHYSICS
COLLEGE PHYSICS
2nd Edition
ISBN: 9781464196393
Author: Freedman
Publisher: MAC HIGHER
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Chapter 24, Problem 57QAP
To determine

(a)

The image formed when a 10.0-cm-tall object is positioned 20.0 cm from the mirror. provide the image distance, the image height, the type of image (real or virtual), and the orientation of the image (upright or inverted).

Expert Solution
Check Mark

Answer to Problem 57QAP

Mirror image appear 6.66 cm behind the mirror
Height the image= 3.33 cm

Virtual, upright image

Explanation of Solution

Given info:

Distance to object from mirror= 20.00 cm

Radius of the curvature= 20.00 cm

Height of the object= 10.00 cm

Formula used:

  f=R2f=focal lengthR=radius of curvature

  1f=1v+1uf=focal lengthv=distance from image to mirroru=distance from object to mirror

  m=vu=HiHom=magnificationv=distance from image to mirrorHi=height of the mirror imageHo=height of the object

Calculations:

  f=R2f=20.00 cm2f=10.0 cm

  1f=1v+1uf=10.00 cmu=20.00 cm110=1v+(120)v=6.66 cm

  vu=HiHo6.66 cm20.00 cm=Hi10.0 cmHi=3.33 cm

Conclusion:

Mirror image appear 6.66 cm behind the mirror
Height the image= 3.33 cm

Virtual, upright image

To determine

(b)

The image formed when a 10.0-cm-tall object is positioned 50.0 cm from the mirror, provide the image distance, the image height, the type of image (real or virtual), and the orientation of the image (upright or inverted).

Expert Solution
Check Mark

Answer to Problem 57QAP

Mirror image appear 8.33 cm behind the mirror
Height the image= 1.67 cm

Virtual, upright image

Explanation of Solution

Given info:

Distance to object from mirror= 50.00 cm

Radius of the curvature= 20.00 cm

Height of the object= 10.00 cm

Formula used:

  f=R2f=focal lengthR=radius of curvature

  1f=1v+1uf=focal lengthv=distance from image to mirroru=distance from object to mirror

  m=vu=HiHom=magnificationv=distance from image to mirrorHi=height of the mirror imageHo=height of the object

Calculations:

  f=R2f=20.00 cm2f=10.0 cm

  1f=1v+1uf=10.00 cmu=50.00 cm110=1v+(150)v=8.33 cm

  vu=HiHo8.33 cm50.00 cm=Hi10.0 cmHi=1.67 cm

Conclusion:

Mirror image appear 8.33 cm behind the mirror
Height the image= 1.67 cm

Virtual, upright image

To determine

(c)

The image formed when a 10.0-cm-tall object is positioned 100.0 cm from the mirror. For each case, provide the image distance, the image height, the type of image (real or virtual), and the orientation of the image (upright or inverted).

Expert Solution
Check Mark

Answer to Problem 57QAP

Mirror image appear 9.10 cm behind the mirror
Height the image= 0.91 cm

Virtual, upright image

Explanation of Solution

Given info:

Distance to object from mirror= 100.00 cm

Radius of the curvature= 20.00 cm

Height of the object= 10.00 cm

Formula used:

  f=R2f=focal lengthR=radius of curvature

  1f=1v+1uf=focal lengthv=distance from image to mirroru=distance from object to mirror

  m=vu=HiHom=magnificationv=distance from image to mirrorHi=height of the mirror imageHo=height of the object

Calculations:

  f=R2f=20.00 cm2f=10.0 cm

  1f=1v+1uf=10.00 cmu=100.00 cm110=1v+(1100)v=9.10 cm

  vu=HiHo9.10 cm100.00 cm=Hi10.0 cmHi=0.91 cm

Conclusion:

Mirror image appear 9.10 cm behind the mirror
Height the image= 0.91 cm

Virtual, upright image

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Chapter 24 Solutions

COLLEGE PHYSICS

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