Converging lenses  (Page 5/5)

The image of an object placed at a distance less than $f$ from the lens is upright. The image is called a virtual image because it is on the same side of the lens as the object and you cannot trace all the light rays directly from the image back to the object.

The image is larger than the object and is located further away from the lens than the object.

The thin lens equation and magnification

The Thin Lens Equation

We can find the position of the image of a lens mathematically as there is a mathematical relation between the object distance, image distance, and focal length. The equation is:

where $f$ is the focal length, $d_o$ is the object distance and $d_i$ is the image distance.

The object distance $d_o$ is the distance from the object to the lens. $d_o$ is positive if the object is on the same side of the lens as the light rays enter the lens. This should make sense: we expect the light rays to travel from the object to the lens. The image distance $d_i$ is the distance from the lens to the image. Unlike mirrors, which reflect light back, lenses refract light through them. We expect to find the image on the same side of the lens as the light leaves the lens. If this is the case, then $d_i$ is positive and the image is real (see [link] ). Sometimes the image will be on the same side of the lens as the light rays enter the lens. Then $d_i$ is negative and the image is virtual ( [link] ). If we know any two of the three quantities above, then we can use the Thin Lens Equation to solve for the third quantity.

Magnification

It is possible to calculate the magnification of an image. The magnification is how much bigger or smaller the image is than the object.

where $m$ is the magnification, $d_o$ is the object distance and $d_i$ is the image distance.

If $d_i$ and $d_o$ are both positive, the magnification is negative. This means the image is inverted, or upside down. If $d_i$ is negative and $d_o$ is positive, then the image is not inverted, or right side up. If the absolute value of the magnification is greater than one , the image is larger than the object. For example, a magnification of -2 means the image is inverted and twice as big as the object.

Converging lenses

1. Which type of lens can be used as a magnifying glass? Draw a diagram to show how it works. An image of the sun is formed at the principal focus of a magnifying glass.
2. In each case state whether a real or virtual image is formed:
   1. Much further than $2f$
   2. Just further than $2f$
   3. At $2f$
   4. Between $2f$ and $f$
   5. At $f$
   6. Between $f$ and 0
   Is a virtual image always inverted?
3. An object stands 50 mm from a lens (focal length 40 mm). Draw an accurate sketch to determine the position of the image. Is it enlarged or shrunk; upright or inverted?
4. Draw a scale diagram (scale: 1 cm = 50 mm) to find the position of the image formed by a convex lens with a focal length of 200 mm. The distance of the object is 100 mm and the size of the object is 50 mm. Determine whether the image is enlarged or shrunk. What is the height of the image? What is the magnification?
5. An object, 20 mm high, is 80 mm from a convex lens with focal length 50 mm. Draw an accurate scale diagram and find the position and size of the image, and hence the ratio between the image size and object size.
6. An object, 50 mm high, is placed 100 mm from a convex lens with a focal length of 150 mm. Construct an accurate ray diagram to determine the nature of the image, the size of the image and the magnification. Check your answer for the magnification by using a calculation.
7. What would happen if you placed the object right at the focus of a converging lens? Hint: Draw the picture.