An object is placed in front of a lens. The image formed by the lens is two times larger than the object.

What kind of lens is it?

- convex (converging)
- concave (diverging)
- it could be either of the above

Diverging mirrors or lenses always produce images smaller than the object. A larger image can only be formed by a converging device, in this case a convex lens.

The focal length of the lens is 10 cm. We want to know where the object is but first consider this. How many solutions are there to the question "What is the object distance?"

- 1
- 2
- more than 2

If the object is between 2F and F we can get a larger real image. If the object is closer to the lens than the focal point we can get a larger virtual image. So, there are two solutions.

First find the solution for the real image:

1/f = 1/d_{i} +1/d_{o}

d_{i} is positive and, from the magnification equation, twice as large as d_{o}. Therefore d_{i} = 2d_{o}

1/f = 1/2d_{o} +1/d_{o} = 3/2d_{o}

Therefore d_{o} = 3f/2 = 15 cm.

d_{i} = 2d_{o} = 30 cm.

Draw the ray diagram to check whether this makes sense.

Now find the solution for the virtual image:

1/f = 1/d_{i} +1/d_{o}

d_{i} is negative and, from the magnification equation d_{i} = -2d_{o}

1/f = -1/2d_{o} +1/d_{o} = 1/2d_{o}

Therefore d_{o} = f/2 = 5 cm.

d_{i} = -2d_{o} = -10 cm.

Draw the ray diagram to check whether this makes sense.