# Image Distance and Magnification

#### erik-the-red

Question

An object is a distance of 23.4 cm from the center of a silvered spherical glass Christmas tree ornament which has a diameter of 5.70 cm.

A What is the position of its image?

B What is the magnification of its image?

I think the object can be treated as a convex mirror.

s (object distance) = 23.4 cm
R = -(5.70)/2 = -2.85 cm

$$\frac{1}{s} + \frac{1}{s'} = \frac{2}{R}$$

Using that equation, I get $$\frac{1}{s'} = \frac{2}{-2.85} - \frac{1}{23.4}$$.

Then, s' = -1.12 cm.

And, $$m = -\frac{s'}{s}$$, so m = .048.

I know for sure that -1.12 cm is not the correct answer, so most likely my second answer is also incorrect.

What did I do wrong?

#### erik-the-red

Did I not provide enough work?

#### Doc Al

Mentor
Your method is fine but check your arithmetic.

#### erik-the-red

Sorry, I still get the same answer.

#### Doc Al

Mentor
Using that equation, I get $$\frac{1}{s'} = \frac{2}{-2.85} - \frac{1}{23.4}$$.

Then, s' = -1.12 cm.
Please redo this calculation.

#### erik-the-red

Doc Al,

Indeed, it was incorrect. Thanks!

### The Physics Forums Way

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving