How to Position an Object for a 2.5x Inverted Image with a Concave Mirror?

[ichilouch]

Homework Statement

Where do you put an object in front of a concave mirror of focal length 10cm to produce an image that is inverted and 2.5 times greater than the object?

Homework Equations

\frac{1}{f}=\frac{1}{P}+\frac{1}{Q}
Where:
Q=Image distance from mirror
P=Object's distance from mirror
f = Focal length

m=\frac{-Q}{P}
m=Magnification

The Attempt at a Solution

m=-2.5
Q=-mP ------ (1)

\frac{1}{f}=\frac{1}{P}+\frac{1}{Q}
\frac{1}{f}=\frac{1}{P}+\frac{1}{-mP}

P=\frac{f}{\frac{m}{m-1}}=\frac{0.1}{\frac{-2.5}{-2.5-1}}
P=0.011 meter

 

[Adithyan]

Focal length of a concave mirror is to be taken negative.

 

[dauto]

Focal length of a concave mirror is to be taken negative.

No it isn't. The most common sign convetion - the Gaussian sign convention - which is used by most elementary optic texts has concave mirrors with positive focal distance and convex mirrors with negative focal distance.

 

[ehild]

P=\frac{f}{\frac{m}{m-1}}=\frac{0.1}{\frac{-2.5}{-2.5-1}}
P=0.011 meter

\frac{0.1}{\frac{-2.5}{-2.5-1}} is not 0.11 m.

ehild

 

[ichilouch]

\frac{0.1}{\frac{-2.5}{-2.5-1}} is not 0.11 m.

ehild

Sorry, Its 0.14m

 

[ehild]

That is correct.

ehild

 

[Adithyan]

No it isn't. The most common sign convetion - the Gaussian sign convention - which is used by most elementary optic texts has concave mirrors with positive focal distance and convex mirrors with negative focal distance.

Oh sorry! I didn't notice that the Poster took object distance as positive. Here in India, we have a different sign convention. According my textbooks, sign of the distance with respect to the pole of the mirror depends on whether the distance is parallel or antiparallel to the incident light ray. If the 'vector' from pole to focus is parallel to the light ray, it is taken positive and negative otherwise.

 

[dauto]

Oh sorry! I didn't notice that the Poster took object distance as positive. Here in India, we have a different sign convention. According my textbooks, sign of the distance with respect to the pole of the mirror depends on whether the distance is parallel or antiparallel to the incident light ray. If the 'vector' from pole to focus is parallel to the light ray, it is taken positive and negative otherwise.

First time a hear of this convention. Could you elaborate?

 

[Adithyan]

First time a hear of this convention. Could you elaborate?

I have posted an image of the sign convention.

 

[dauto]

I have posted an image of the sign convention.

That seems to be the Cartesian sign convention which is the second most common sign convention as far as I know. I thought you were talking about some other convention.