Image Distance of a concave mirror ( with algebra)

In summary: Thanks!In summary, the problem involves a concave mirror with a focal length of 34.4 cm and an image located at one-third the distance of the object from the mirror. Using the mirror equation (1/f)=(1/do)+(1/di), we can substitute 1/3(do) for Di to get (1/f)=(1/do)+[1/(1/3)do]. Solving for do, we get do=4f, which is the correct solution.
  • #1
Cheezay
26
0

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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  • #2
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 ?
 
  • #3
Yes, that's how it works out. But solving for do, i get a wrong answer. Perhaps i have my equation set up wrong?
 
  • #4
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
 
  • #5
(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
 
  • #6
Thanks v_bachtiar, that helped a lot!
 
  • #7
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.
 

Related to Image Distance of a concave mirror ( with algebra)

1. What is the formula for calculating the image distance of a concave mirror?

The formula for calculating the image distance of a concave mirror is d = f * (1 + 1/m), where d is the image distance, f is the focal length of the mirror, and m is the object distance.

2. How do you determine the focal length of a concave mirror?

The focal length of a concave mirror can be determined by using the formula f = R/2, where R is the radius of curvature of the mirror.

3. How does the position of the object affect the image distance in a concave mirror?

The position of the object relative to the focal point of the concave mirror affects the image distance. When the object is placed between the focal point and the mirror, the image distance is positive and the image is virtual. When the object is placed beyond the focal point, the image distance is negative and the image is real.

4. Can the image distance of a concave mirror be greater than the object distance?

Yes, the image distance of a concave mirror can be greater than the object distance. This occurs when the object is placed beyond the center of curvature of the mirror.

5. How does the size of the object affect the image distance in a concave mirror?

The size of the object has no direct effect on the image distance in a concave mirror. However, the size of the image is affected by the size of the object and the ratio of the image distance to the object distance, also known as the magnification.

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