# Archived Angular magnification and the magnifying glass

1. Feb 17, 2013

### Reservist

1. The problem statement, all variables and given/known data

An engraver uses a magnifying glass (f=9.5 cm) to examine some work. The image he sees is located 25 cm from his eye, which is his near point. A) what is the distance between the work and the magnifying glass? b) what is the angular magnification of the magnifying glass?

2. Relevant equations

M= (1/f -1/di)N
1/di + 1/do = 1/f

3. The attempt at a solution

A) 1/25 + 1/do= 1/9.5 do= 15.3 cm

b) m=(1/9.5 - 1/25) 25 M= 1.63

Can some one check/ correct my work, not sure if I'm using the correct formulas. Thank you.

2. Oct 29, 2016

### Ibix

This is a virtual image. Using the sign convention you have, the image distance is negative. So $$\begin {eqnarray*} \frac 1{d_o}&=&\frac 1f-\frac 1{d_i}\\ &=&\frac 1 {9.5}-\frac 1 {-25}\end {eqnarray*}$$I make that $d_o=6.88$cm for part (a).

Angular magnification is just $M=-d_i/d_o$. Not sure what your formula is since you don't define N. In this case I make it 25/6.88=3.63 for part (b).