Solve Concave Mirror Problems: Find Image Distance & Focal Length

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SUMMARY

The discussion focuses on solving problems related to concave mirrors, specifically calculating image distance and focal length. A concave mirror produces a virtual image that is three times taller than the object, which is positioned 21 cm in front of the mirror. The correct image distance is calculated using the formula di/do = 3, leading to an image distance (di) of -7.0 cm. The focal length can be determined using the mirror formula 1/f = 1/do + 1/di once the correct image distance is established.

PREREQUISITES
  • Understanding of concave mirrors and their properties
  • Familiarity with the mirror formula: 1/f = 1/do + 1/di
  • Knowledge of magnification equations: hi/ho and di/do
  • Ability to differentiate between real and virtual images
NEXT STEPS
  • Review the mirror formula and its application to concave mirrors
  • Study the concept of magnification in optics
  • Explore the differences between real and virtual images in concave mirrors
  • Practice solving additional problems involving concave mirrors and image formation
USEFUL FOR

Students studying optics, physics educators, and anyone seeking to understand the principles of concave mirrors and image formation.

matt72lsu
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Homework Statement


1)A concave mirror produces a virtual image that is three times as tall as the object.
a) If the object is 21 {\rm cm} in front of the mirror, what is the image distance?
b) What is the focal length of this mirror?

Homework Equations


hi/ho
di/do
1/f= 1/do + 1/di


The Attempt at a Solution


hi/ho = 3 thus do/di = 3 a) di = -do /3 = -7.0cm (was wrong)
thus i could not do the second part. Where did I go wrong? And if it was a real image instead of a virtual image, what what you do different? Thanks
 
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Hello matt72lsu,

matt72lsu said:

Homework Statement


1)A concave mirror produces a virtual image that is three times as tall as the object.
a) If the object is 21 {\rm cm} in front of the mirror, what is the image distance?
b) What is the focal length of this mirror?

Homework Equations


hi/ho
di/do
1/f= 1/do + 1/di

hi/ho and do/do are not really equations. If you put these in full equation form, ensuring that they apply to virtual images in concave mirrors, it might help.

The Attempt at a Solution


hi/ho = 3 thus do/di = 3 a) di = -do /3 = -7.0cm (was wrong)

I believe you've got your do and di switched around.

thus i could not do the second part. Where did I go wrong? And if it was a real image instead of a virtual image, what what you do different? Thanks

For one thing, knowing whether the image is real or virtual gives you information about where the object is relative to the mirror's focal length (is the object between the focal length and the mirror or is it on the far side of the focal length?).
 
i'm sorry, i just have no clue what to do. the whole 3 times larger part is throwing me off
 
matt72lsu said:
i'm sorry, i just have no clue what to do. the whole 3 times larger part is throwing me off

If the image is 3 times as large, than it's 3 times as far away. :wink:

Start by going back to your original equations.

Earlier you calculated
hi/ho = 3 thus do/di = 3

But that should be di/do = 3.
 
i got this one already
 

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