Concave Spherical Mirrors: Object Position for Inverted and Enlarged Image?

[roam]

Homework Statement

A concave spherical mirror has a radius of curvature of magnitude 27.1 cm. Determine the object position for which the resulting image is inverted and larger than the object by a factor of 4.00.

Homework Equations

Mirror equation in terms of focal length: 1/p + 1/q = 1/f

The Attempt at a Solution

I'm not getting the correct answer for this question. Here's what I've done so far:

First we find the focal length

[tex]f=\frac{R}{2} \implies \frac{27.1}{2}=13.55 \ cm[/tex]

[tex]M = \frac{-q}{p}=4 \implies q=-4p[/tex]

Substituting in

[tex]\frac{1}{p} - \frac{1}{4p} = \frac{1}{13.55}[/tex]

[tex]\frac{3}{4p} = \frac{1}{13.55} \implies p =10.162[/tex]

I don't know why the answer is wrong. I know that image is inverted when Magnification is negative. Here it is not -ve because the questions says "larger" so I took it as positive...

 

[PeterO]

Homework Statement

A concave spherical mirror has a radius of curvature of magnitude 27.1 cm.

Homework Equations

Mirror equation in terms of focal length: 1/p + 1/q = 1/f

The Attempt at a Solution

I'm not getting the correct answer for this question. Here's what I've done so far:

First we find the focal length

[tex]f=\frac{R}{2} \implies \frac{27.1}{2}=13.55 \ cm[/tex]

[tex]M = \frac{-q}{p}=4 \implies q=-4p[/tex]

Substituting in

[tex]\frac{1}{p} - \frac{1}{4p} = \frac{1}{13.55}[/tex]

[tex]\frac{3}{4p} = \frac{1}{13.55} \implies p =10.162[/tex]

I don't know why the answer is wrong. I know that image is inverted when Magnification is negative. Here it is not -ve because the questions says "larger" so I took it as positive...

What are you trying to find? You haven't said.

 

[roam]

What are you trying to find? You haven't said.

Oops. The question I was trying to ask was:

Determine the object position for which the resulting image is inverted and larger than the object by a factor of 4.00.

 

[PeterO]

Oops. The question I was trying to ask was:

Determine the object position for which the resulting image is inverted and larger than the object by a factor of 4.00.

You are having trouble with the minus signs.

There are two conventions out there one says magnification is -p/q the other says it is p/q

one of them has 1/p + 1/q = 1/r , the other has 1/p - 1/q = 1/r

make everything positive, and work on a positive magnification meaning inverted.

 

[roam]

You are having trouble with the minus signs.

There are two conventions out there one says magnification is -p/q the other says it is p/q

one of them has 1/p + 1/q = 1/r , the other has 1/p - 1/q = 1/r

make everything positive, and work on a positive magnification meaning inverted.

So, are you saying I should have used M=q/p so that q=4p? Because I am using the +ve equation 1/p+1/q=1/f.

 

[PeterO]

So, are you saying I should have used M=p/q?

Try it and see.

 

[roam]

YES! It worked. Thanks for clarifying this, I REALLY appreciate all your help.