Solve Concave Mirror Problems: Find Image Distance & Focal Length

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A concave mirror produces a virtual image three times taller than the object, which is placed 21 cm in front of the mirror. The initial attempt to find the image distance resulted in an incorrect calculation of -7.0 cm. The correct relationship between object distance (do) and image distance (di) should be di/do = 3, indicating that the image distance is actually three times the object distance. Understanding whether the image is real or virtual is crucial for determining the object's position relative to the mirror's focal length. Clarifying these relationships and equations will help solve the problem correctly.
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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
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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