Image Distance of a concave mirror ( with algebra)

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Homework Help Overview

The discussion revolves around a problem involving a concave mirror with a specified focal length, where participants are tasked with calculating the object distance based on the relationship between the image distance and object distance.

Discussion Character

  • Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of the mirror equation and the substitution of image distance in terms of object distance. There are attempts to manipulate the equation algebraically to isolate the object distance.

Discussion Status

Some participants have provided guidance on the algebraic manipulation of the equation, while others express confusion about their calculations. There is an acknowledgment of a potential error in the setup of the equation, and clarification is sought regarding the correct interpretation of terms.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can share or the methods they can use. There is a focus on ensuring the correct application of the mirror equation and the relationships between the distances involved.

Cheezay
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Homework Statement


A concave mirror with a focal length of 34.4 cm produces an image whose distance from the mirror is one-third the object distance. Calculate the object distance.


Homework Equations


(1/f)=(1/do) + (1/di)


The Attempt at a Solution


I know that i have to use the mirror equation (posted above). I substitute 1/3(do) in for Di, because the problems states that "... produces an image whose distance from the mirror is one-third the object distance" so my equation now looks like this: (1/f)=(1/do) + [1/(1/3)do].
My problem, sad as it seems, comes during the algebra work. I can't for the life of me get the do terms by themselves to solve. Could someone please help refresh my memory? Basically what I've been doing is adding the do terms to get (1 and 1/3)do = .344m and then divide 1 by the answer that i get solving for do. Thanks for any help you can give!
 
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Cheezay said:

Homework Statement


A concave mirror with a focal length of 34.4 cm produces an image whose distance from the mirror is one-third the object distance. Calculate the object distance.

Homework Equations


(1/f)=(1/do) + (1/di)

The Attempt at a Solution


I know that i have to use the mirror equation (posted above). I substitute 1/3(do) in for Di, because the problems states that "... produces an image whose distance from the mirror is one-third the object distance" so my equation now looks like this: (1/f)=(1/do) + [1/(1/3)do].
My problem, sad as it seems, comes during the algebra work. I can't for the life of me get the do terms by themselves to solve. Could someone please help refresh my memory? Basically what I've been doing is adding the do terms to get (1 and 1/3)do = .344m and then divide 1 by the answer that i get solving for do. Thanks for any help you can give!

Doesn't it work out from your equation that 4/3*f = do ?
 
Yes, that's how it works out. But solving for do, i get a wrong answer. Perhaps i have my equation set up wrong?
 
Cheezay said:
Yes, that's how it works out. But solving for do, i get a wrong answer. Perhaps i have my equation set up wrong?

To be clear, if you arrived at
(1 and 1/3)do = .344m
this is incorrect. It should be 3/4*do = .344
 
(1/f)=(1/do)+(1/di)
(1/f)=(1/do)+(1/((1/3)do))
(1/f)=(1/do)+(1/(do/3))
(1/f)=(1/do)+(3/do)
(1/f)=(4/do)
do=4f

remember:
[1/(1/3)do]
is equal to 1/(do/3)=3/do
 
Thanks v_bachtiar, that helped a lot!
 
v_bachtiar said:
remember:
is equal to 1/(do/3)=3/do

Yikes. You're right. I took it as 1/(3*do) which is wrong. Good catch.
 

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