# A Pair of Lens Problems

1. Mar 25, 2005

### evilempire

1. A coin is located 19.6cm to the left of a converging lens
(f = 16.5cm). A second, identical lens is placed to the right of the first lens, such that the image formed by the combination has the same size and orientation as the original coin. Calculate the separation between the lenses
.

This one I'm pretty lost on. I know that 1/do + 1/di = 1/f, but the 'same size and orientation' is particularly tripping me up.

2. A person holds a book 27.7cm in front of the effective lens of her eye; the print in the book is 1.80mm high. If the effective lens of the eye is located 1.75cm from the retina, what is the size (including the sign) of the print image on the retina?

I thought the equation to solve this was hi/ho=-(di/do), but I'm apparantly mistaken as using this equation produces an incorrect result. Of course, I could have misused it.

Any help is appreciated, and thanks.

2. Mar 26, 2005

### Andrew Mason

I think your approach is right. The focal length of the eye adjusts so that the image is in focus on the retina. So that gives you the magnification factor:

$$M = - \frac{i}{o} = - 1.75/27.7 = - .063$$

So image height in the eye is: M*1.8 = .063*1.8 = -.11 mm. (inverted)

I'll have to think about the first one.

AM

3. Mar 28, 2005

### Andrew Mason

I am a little rusty on optics but I think I have this figured out.

1. find the image distance from the first lens using the lens equation.

2. find the magnification using M1 = -i/o where o is 19.5 and i is from 1.

3. find the magnification factor (reduction) needed to reduce the image back to the original height (ie. M2 = -1/M1). This gives you the ratio i2/o2

4. find the new object distance using the lens equation and replacing i2 with (-1/M1)o2.

5. add i1 and o2 to get the total separation.

I get 214.5 cm.

AM