 # 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 (ho) = 5 cm

The object distance (do) = 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 (di) :

1/di = 1/f – 1/do = 1/15 – 1/5 = 1/15 – 3/15 = -2/15

di = -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 = – di / do = -(-7.5)/5 = 7.5/5 = 1.5

The image height (hi) :

m = hi / ho

hi = m ho = (1.5)5 = 10/3 = 7.5 cm

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 (di) :

1/di = 1/f – 1/do = 1/15 – 1/30 = 2/30 – 1/30 = 1/30

di = 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 = – di / do = -(30)/30 = -30/30 = -1

The image height (hi) :

m = hi / ho

hi = m ho = (-1)10 = -10 cm

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 (do) = 30 cm

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

Solution :

The image distance (di) :

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

di = 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 = – di / do = -(60)/30 = -2

The image is 2 times greater than the object.

The image height (hi) :

m = hi / ho

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

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

See also  Partially inelastic collisions in one dimension - problems and solutions

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

Known : Object distance (do) = 20 cm

Image distance (di) = 30 cm

Wanted : The focal length (f)

Solution :

1/do + 1/di = 1/f

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

di = 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 (do) = 30 cm

Wanted : The magnification of image (M)

Solution :

Image distance :

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

di = 60/1 = 60 cm

The magnification of image :

M = di / do = 60 cm / 30 cm = 60/30 = 2 times

[wpdm_package id=’864′]