Solving Inverted Image w/Concave Mirror: 22 cm, 44 cm, & Focal Length

In summary, the problem states that an inverted image is magnified by 2 when the object is placed 22 cm in front of a concave mirror. The task is to determine the image distance and focal length of the mirror. Using the formula M = -di/do, the image distance is calculated to be -44 cm. When this value is substituted into the equation 1/f = 1/do + 1/di, the focal length is found to be 44 cm. The negative sign in this equation is a convention used to indicate an inverted image. By convention, a positive magnification indicates an upright image, while a negative magnification indicates an inverted image. This is consistent with the given problem, as an inverted image
  • #1
noobie!
58
0

Homework Statement



An inverted image is magnified by 2 when the object is placed 22 cm in front of a concave mirror. Determine the image distance and the focal length of the mirror.

Homework Equations



don't have,sorry

The Attempt at a Solution


firstly,i use M= -di/do then i got di=-44cm ,then substitute in into the equation 1/f=1/do + 1/di ..1/f= 1/22 + (- 1/44) =44cm..did i get my steps correct?then i refer to answers,i noticed that their solution is 1/f= 1/22cm + 1/44cm without the negative..why could it be so?so,are my steps correct?the solution also stated image is real if its being describe upright..according to what I've read,it should be when a real image will describe as inverted image while virtual is to be described as upright and located behind the mirror,its so contradicting..so could you please help me to clear my doubts..thanks for your kind help!:smile:
 
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  • #2
  • #3
LowlyPion said:
http://en.wikipedia.org/wiki/Concave_mirror#Analysis

I think by the statement of the problem your magnification is -2 to begin with.

Also you will be dealing with this case:
http://en.wikipedia.org/wiki/File:Concavemirror_raydiagram_2FE.svg

oh..did you actually mean that convention merely a choice for me..so,if i place a either a negative sign or i don't will also be correct?..then the problem the magnification should be -2 instead of 2?if yes then i understand completely...?
 
  • #4
noobie! said:
oh..did you actually mean that convention merely a choice for me..so,if i place a either a negative sign or i don't will also be correct?..then the problem the magnification should be -2 instead of 2?if yes then i understand completely...?

The sign is apparently a convention that indicates the case that you have.

What they gave you was that the magnification is 2. But inverted ... so it's -2.

The formula is m = - di/do = -2 ... so the signs cancel looks like to me.

2 = di/do
 
  • #5
LowlyPion said:
The sign is apparently a convention that indicates the case that you have.

What they gave you was that the magnification is 2. But inverted ... so it's -2.

The formula is m = - di/do = -2 ... so the signs cancel looks like to me.

2 = di/do

ah..i got your point..understood..thanks for your kind help..thanks!:smile:
 

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

The focal length of a concave mirror can be determined by measuring the distance between the mirror and the point where the reflected light rays converge, known as the focal point. This distance is equal to half the radius of curvature of the mirror. Alternatively, the focal length can also be calculated using the equation 1/f = 1/do + 1/di, where f is the focal length, do is the object distance, and di is the image distance.

2. What is the difference between a virtual image and a real image?

A virtual image is an image that appears to be behind the mirror and cannot be projected onto a screen. It is formed by the apparent intersection of reflected light rays. A real image, on the other hand, is formed in front of the mirror and can be projected onto a screen. It is formed by the actual intersection of reflected light rays.

3. How do you determine the position of an inverted image formed by a concave mirror?

The position of an inverted image formed by a concave mirror can be determined by measuring the distance between the mirror and the object, known as the object distance, and the distance between the mirror and the image, known as the image distance. These distances can then be used to calculate the magnification of the image, which can be used to determine the position of the image relative to the mirror.

4. Can the image formed by a concave mirror be larger than the object?

Yes, the image formed by a concave mirror can be larger than the object. This occurs when the object is placed between the focal point and the mirror. In this case, the image will be magnified and appear larger than the object. However, the image will still be inverted.

5. How does the distance between the object and the mirror affect the image formed by a concave mirror?

The distance between the object and the mirror, also known as the object distance, affects the size and orientation of the image formed by a concave mirror. As the object distance increases, the image will become smaller and closer to the focal point. When the object is placed at the focal point, the image will become infinitely large. When the object is placed beyond the focal point, the image will become inverted and move further away from the mirror.

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