Converging lens – problems and solutions

1. A 5-cm high object is placed 5 cm from a 15-cm focal length converging lens. Determine the image distance, the magnification of the image, the image height and the properties of the image.

Known :

The focal length (f) = 15 cm

The plus sign indicates that the focal point is real or the rays pass through the point.

The object height (h_o) = 5 cm

The object distance (d_o) = 5 cm

Wanted : the image distance, the magnification of image, the image height and the properties of image

Solution :

Formation of image by converging lens :

The image distance (d_i) :

1/d_i = 1/f – 1/d_o = 1/15 – 1/5 = 1/15 – 3/15 = -2/15

d_i = -15/2 = -7.5 cm

The minus sign indicates that the image is virtual or the rays do not pass through the image.

The magnification of image (m) :

m = – d_i / d_o = -(-7.5)/5 = 7.5/5 = 1.5

The plus sign indicates that the image is upright.

The image height (h_i) :

m = h_i / h_o

h_i = m h_o = (1.5)5 = 10/3 = 7.5 cm

The plus sign indicates that the image is upright.

The properties of the image

– virtual

– upright

– The image greater than the object

– The image distance is greater than the object distance

2. A 10-cm high object is placed 30 cm from a 15-cm focal length converging lens. Determine the image distance, the magnification of the image, the image height and the properties of the image.

Known :

The focal length (f) = 15 cm

The plus sign indicates that the focal point is real or the rays pass through the point.

The object height (h) = 10 cm

The object distance (s) = 30 cm

Wanted: The image distance, the magnification of the image, the image height and the properties of the image

Solution :

Formation of the image by the converging lens :

The image distance (d_i) :

1/d_i = 1/f – 1/d_o = 1/15 – 1/30 = 2/30 – 1/30 = 1/30

d_i = 30/1 = 30 cm

The plus sign indicates that the image is real or the rays pass through the image.

The magnification of image (m) :

m = – d_i / d_o = -(30)/30 = -30/30 = -1

The minus sign indicates that the image is inverted.

The image height (h_i) :

m = h_i / h_o

h_i = m h_o = (-1)10 = -10 cm

The minus sign indicates that the image is inverted.

The properties of the image

– real or the rays pass through the image

– inverted

– The image height is the same as the object height.

– The image distance is the same as the object distance.

3. A 10-cm high object is placed 30 cm from a 20-cm focal length converging lens. Determine the image distance, the magnification of the image, the image height and the properties of the image.

Known :

The focal length f) = 20 cm

The plus sign indicates that the focal point is real or the rays pass through the point.

The object height (h) = 10 cm

The object distance (d_o) = 30 cm

Wanted : The image distance, the magnification of image, the image height and the properties of image

Solution :

The image distance (d_i) :

1/d_i = 1/f – 1/d_o = 1/20 – 1/30 = 3/60 – 2/60 = 1/60

d_i = 60/1 = 60 cm

The plus sign indicates that the image is real or the rays pass through the image.

The magnification of image (m) :

m = – d_i / d_o = -(60)/30 = -2

The minus sign indicates that the image is inverted.

The image is 2 times greater than the object.

The image height (h_i) :

m = h_i / h_o

hi = m h_o = (-2)10 = -20 cm

The minus sign indicates that the image is inverted.

The properties of the image

– real

– inverted

– the image height is greater than the object height

– the image distance is greater than the object distance

4. Based on the figure below, determine the focal length of the converging lens!

Known :

Object distance (d_o) = 20 cm

Image distance (d_i) = 30 cm

Wanted : The focal length (f)

Solution :

1/d_o + 1/d_i = 1/f

d_o = object distance (plus sign because rays pass through object)

d_i = image distance (plus sign because rays pass through image)

f = focal length (plus sign because rays pass through the focal point or the focal point is real)

The focal length :

1/f = 1/20 + 1/30 = 3/60 + 2/60 = 5/60

f = 60/5 = 12 cm

5. Object’s distance is 30 cm and the focal length is 20 cm. Determine the magnification of the image.

Known :

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

The plus sign indicates that the focal point is real or rays pass through the image.

Object distance (d_o) = 30 cm

Wanted : The magnification of image (M)

Solution :

Image distance :

1/d_i = 1/f – 1/do = 1/20 – 1/30 = 3/60 – 2/60 = 1/60

d_i = 60/1 = 60 cm

The magnification of image :

M = d_i / d_o = 60 cm / 30 cm = 60/30 = 2 times

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