Basic Physics

# Image formation by diverging lens (concave lens)

To understand the image formation by the concave lens, look at the following example problems and solution. In this case, the object is assumed to be at a certain distance from the concave lens, then draw the image formation by the concave lens, the image distance from the concave lens and the magnification of the image by the concave lens.

The focal length of a concave lens (diverging lens) is 20 cm. Objects with a height of 10 cm are located at the left of the lens. Determine the image distance, the magnification of the image and the height of the image, if:

a) the object distance is smaller than the focal length of the concave lens

b) the object distance is the same as the focal length of the concave lens

c) the object distance is greater than the focal length of the concave lens

Known:

The focal length of the concave lens (f) = -20 cm

The focal length of the concave lens has a negative sign because the focal point of the concave lens is virtual, where the light beam does not pass through the focal point.

The height of object (h) = 10 cm

Solution:

The object distance is smaller than the focal length of the concave lens (do > f)

Suppose the object distance is 5 cm, 10 cm and 15 cm.

a) The focal length is -20 cm and if the object distance (do) = 5 cm The image distance (di)

1/di = 1/f – 1/do = -1/20 – 1/5 = -1/20 – 4/20 = -5/20

di = -20/5 = -4 cm

The image distance sign negative, means the image is virtual, where the beam of light does not pass through the image.

The image distance 4 cm, smaller than the object distance, 5 cm.

The magnification of image (M)

M = -di / do = -(-4)/5 = 4/5 = 0.8

The magnification of image smaller than 1 means that the image is reduced or the image size is smaller than the object size.

The height of image (h’)

The equation of the magnification of image :

M = h’ / h

h’ = M h = (0.8)(10 cm) = 8 cm

The height of image is sign positive, means that the image is upright.

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b) The focal length is -20 cm and if the object distance (do) = 10 cm The image distance (di)

1/di = 1/f – 1/do = -1/20 – 1/10 = -1/20 – 2/20 = -3/20

di = -20/3 = -6.7 cm

The image distance is negative, means that the image is virtual, where the beam of light does not pass through the image.

The image distance 6.7 cm, smaller than the object distance, 10 cm.

The magnification of image (M)

M = -di / do = -(-6.7)/10 = 6.7 / 10 = 0.67

The magnification of image is smaller than 1 means that the image is reduced or the image size is smaller than the object size.

The height of image (h’)

The equation of the magnification of image:

M = h’/h

h’ = M h = (0.67)(10 cm) = 6.7 cm

The height of image is sign positive, means that the image is upright.

c) The focal length is -20 cm and the object distance (do) = 15 cm The image distance (di)

1/di = 1/f – 1/do = -1/20 – 1/15 = -3/60 – 4/60 = -7/60

di = -60/7 = -8.6 cm

The image distance is negative, means that the image is virtual, where the beam of light does not pass through the image.

The image distance is 8.7 cm, smaller than the object distance 15 cm.

The magnification of image (M)

M = -di / do = -(-8.6)/10 = 8.6 / 10 = 0.86

The magnification of image smaller than 1 means that the image size is smaller than the object size.

The height of image (h’)

The equation of the magnification of image:

M = h’/h

h’ = M h = (0.86)(10 cm) = 8.6 cm

The height of image is positive, means that the image is upright.

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The object distance is the same as the focal length of the concave lens (do = f)

The focal length (f) = the object distance (do) = 20 cm. The image distance (di)

1/di = 1/f – 1/do = -1/20 – 1/20 = -2/20 = -2/20

di = -20/2 = -10 cm

The image distance is negative, means that the image is virtual, where the beam of light does not pass through the image.

The image distance is 10 cm, smaller than the object distance, 20 cm.

The magnification of image (M)

M = -di / do = -(-10)/20 = 10/20 = 0.5

The magnification of image is smaller than 1, means that the image size is smaller than the object size.

The height of the image (h’)

The equation of the magnification of the image :

M = h’/h

h’ = M h = (0.5)(10 cm) = 5 cm

The height of the image is positive, means that the image is upright.

The object distance is greater than the focal length of the concave lens (do > f)

Suppose the distance of objects is 30 cm and 40.

a) The focal length is -20 cm and if the object distance (do) = 30 cm The image distance (di)

1/di = 1/f – 1/do = -1/20 – 1/30 = -3/60 – 2/60 = -5/60

di = -60/5 = -12 cm

The image distance is negative, means that the image is virtual, where the beam of light does not pass through the image.

The image distance is 12 cm, smaller than the object distance 30 cm.

The magnification of image (M)

M = -di / do = -(-12)/30 = 12/30 = 0.4

The magnification of image smaller than 1, means that the image size is smaller than the object size.

The height of image (h’)

The equation of the magnification of image:

M = h’/h

h’ = M h = (0.4)(10 cm) = 4 cm

The height of image is sign positive, means that image is upright.

b) The focal length -20 cm and if the object distance (do) = 40 cm The image distance (di)

1/di = 1/f – 1/do = -1/20 – 1/40 = -2/40 – 1/40 = -3/40

di = -40/3 = -13 cm

The image distance is negative, means that the image is virtual, where the beam of light does not pass through the image.

The image distance is 13 cm, smaller than the object distance 40 cm.

The magnification of image (M)

M = -di / do = -(-13)/40 = 13/40 = 0.3

The magnification of image smaller than 1, means that the image size is smaller than the object size.

The height of the image (h’)

The equation of the magnification of the image:

M = h’/h

h’ = M h = (0.3)(10 cm) = 3 cm

The height of the image is signed positive, means that the image is upright.