# Homework Help: Near Point Distance of 25cm, how far should object be?

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1. Mar 11, 2016

### romeIAM

1. The problem statement, all variables and given/known data
A magnifier has a magnification of

How far from the lens should an object be held so that its image is seen at the near-point distance of 25 cm?
Assume that your eye is immediately behind the lens.

Express your answer to two significant figures and include the appropriate units.

2. Relevant equations
M = Di/Do

M = 9
Di = 25 cm
Do = ?

3. The attempt at a solution
I thought this problem would be straight forward. Just solve for Do so I ended up with Do = Di / M
Do = 25 cm / 9 which i i though equals 2.8 cm. But it is wrong. Where am i messing up?

2. Mar 11, 2016

### ProfuselyQuarky

No, it's 0.36. Based on your equation . . .

M = Di / Do
9 = 25 / Do
9/25 = Do
0.36 = Do

Just a small math error. You should remember to check your work. Instead of Do = Di / M, it is Do = M / Di

3. Mar 11, 2016

### Staff: Mentor

The second equation would be 9/25 = 1/Do.
Easy to see if you work with units, your result has inverse centimeters instead of centimeters, it cannot be right.

How does the magnifier work if the eye is directly behind the lens? Do you have a sketch?

4. Mar 11, 2016

### Merlin3189

This makes no sense at all to me.

As far as the OP is concerned, I agree with that calculation. If it is the wrong answer, I can only guess that 9x might refer to the area magnification rather than the linear magnification.

5. Mar 11, 2016

### ProfuselyQuarky

Ah, okay, I see where I went wrong. I see where the OP is confused now, but now I'm confused, too.
I should really listen to my own reproach. My apologies; I've had a really awful week.

Okay, so now I agree with the OP's calculations. I don't know why it's wrong.

6. Mar 11, 2016

### Merlin3189

I'm always putting my foot in it, especially with errors like that. It's often easier to see other's errors than your own.

Having looked a bit further into this, I see there is a further possibility. The idea of area magnification does not seem to be common, so I'll leave that for now. But the magnification achieved by a lens does depend on where you hold the object and form the image.
The maximum mag is achieved when the image is at the near point and the object is nearer than f. That is the situation we are looking at here.
But "normal adjustment" of an optical instrument assumes the image is at ∞ and the object at f. This is supposed to be the most comfortable position, but gives a lower magnification.

Maybe, they are defining magnification at normal adjustment and asking us to find out the focal length, then use this to calculate the object distance for maximum magnification. (There is actually a simpler procedure, since it turns out that MNP = M + 1)
The magnification in normal adjustment, when the object is held at f, is
$M = \frac{\ angle\ subtended\ by\ the\ image\ }{\ angle\ subtended\ by\ object\ at\ closest\ viewing\ point\ }$

$M = \frac{\ the\ angle\ subtended\ by\ the\ object\ when\ at\ f\ }{\ the\ angle\ subtended\ by\ the\ object\ when\ at\ the\ NP\ }$

$M = \frac{ h}{f}$ / $\frac{h}{25cm}$ = $\frac{25cm}{f}$

Since this is a HW Q, I'll pause there. You can look these formulae up on Hyperphysics or somewhere, or ask for further explanation as you choose.