# Diverging Lens

1. Jul 31, 2011

### fromthepast

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

A lens forms an image of an object. The object is 16 cm from the lens. The image is 12 cm
from the lens on the same side as the object. (a) What is the focal length of the lens? Is the
lens converging or diverging? (b) If the object is 8.5 mm tall, how tall is the image? Is it erect or inverted? Draw a principle-ray diagram.

2. Relevant equations

1/f = 1/s + 1/s'

3. The attempt at a solution

I got 6.8cm for the focal length and .638cm for the height. When I draw my principle-ray diagram, the object and image are to the left of the second focal point, which is to the left of the lens. Shouldn't the 2nd focal point be between the object and image on the left side of the lens?

2. Jul 31, 2011

### fromthepast

Anybody?

3. Jul 31, 2011

### ehild

The image is at the same side as the object. Is the image real or virtual?

ehild

4. Jul 31, 2011

### fromthepast

virtual

5. Jul 31, 2011

### ehild

Your result for the focal length is not correct. The virtual image distance has minus sign in the lens formula. Calculate again.

ehild

6. Jul 31, 2011

### fromthepast

-48cm?

7. Jul 31, 2011

### ehild

Yes, the focal length is -48 cm. The lens is diverging so the focal length is negative.

ehild