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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:

1 f = 1 d o + 1 d i

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.

m = - d i d o

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.

An object is placed 6 cm from a converging lens with a focal point of 4 cm.

  1. Calculate the position of the image
  2. Calculate the magnification of the lens
  3. Identify three properties of the image
  1. f = 4 cm d o = 6 cm d i = ? m = ?

    Properties of the image are required.

  2. 1 f = 1 d o + 1 d i 1 4 = 1 6 + 1 d i 1 4 - 1 6 = 1 d i 3 - 2 12 = 1 d i d i = 12 cm
  3. m = - d i d o = - 12 6 = - 2
  4. The image is real, d i is positive, inverted (because the magnification is negative) and enlarged (magnification is > 1)

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An object is placed 5 cm to the left of a converging lens which has a focal length of 2,5 cm.

  1. What is the position of the image?
  2. Is the image real or virtual?
  1. Draw the lens, the object and mark the focal points.

    • R 1 goes from the top of the object parallel to the principal axis, through the lens and through the focal point F 2 on the other side of the lens.
    • R 2 goes from the top of the object through the focal point F 1 , through the lens and out parallel to the principal axis.
    • R 3 goes from the top of the object through the optical centre with its direction unchanged.

  2. The image is at the place where all the rays intersect. Draw the image.

  3. The image is 5 cm away from the lens, on the opposite side of the lens to the object.

  4. Since the image is on the opposite side of the lens to the object, the image is real.

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An object, 1 cm high, is placed 2 cm to the left of a converging lens which has a focal length of 3,0 cm. The image is found also on the left side of the lens.

  1. Is the image real or virtual?
  2. What is the position and height of the image?
  1. Draw the lens, principal axis, focal points and the object.

    • R 1 goes from the top of the object parallel to the principal axis, through the lens and through the focal point F 2 on the other side of the lens.
    • R 2 is the light ray which should go through the focal point F 1 but the object is placed after the focal point! This is not a problem, just trace the line from the focal point F 1 , through the top of the object, to the lens. This ray then leaves the lens parallel to the principal axis.
    • R 3 goes from the top of the object through the optical centre with its direction unchanged.
    • Do not write R 1 , R 2 and R 3 on your diagram, otherwise it becomes too cluttered.
    • Since the rays do not intersect on the right side of the lens, we need to trace them backwards to find the place where they do come together (these are the light gray lines). Again, this is the position of the image.

  2. The image is 6 cm away from the lens, on the same side as the object.

  3. The image is 3 cm high.

  4. Since the image is on the same side of the lens as the object, the image is virtual.

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Phet simulation for geometrical optics

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 2 f
    2. Just further than 2 f
    3. At 2 f
    4. Between 2 f 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.

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Source:  OpenStax, Siyavula textbooks: grade 11 physical science. OpenStax CNX. Jul 29, 2011 Download for free at http://cnx.org/content/col11241/1.2
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