Basic Physics

# Diverging lens – problems and solutions

1. A 5-cm high object is placed 15 cm from a 30-cm focal length diverging lens. Determine the image distance, the magnification of the image, the image height, and properties of the image.

Known :

The focal length (f) = -30 cm

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

The object height (ho) = 5 cm

The object distance (do) = 15 cm

Wanted: The image distance (di), the magnification of image (m), the image height (hi) and the properties of the image.

Solution :

Formation of image by diverging lens : The image distance (di) :

1/di = 1/f – 1/do = -1/30 – 1/15 = -1/30 – 2/30 = -3/30

di = -30/3 = -10 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 = -(-10)/15 = 10/15 = 2/3

The image 2/3 smaller than the object.

The image height (hi) :

m = hi / ho

hi = m ho = (2/3)5 = 10/3 = 3.3 cm

The properties of the image :

The properties of the image formed by a diverging mirror :

– virtual

– upright

– the image smaller than the object

– the image distance smaller than the object distance

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2. A 10-cm high object is placed 60 cm from a 30-cm focal length diverging lens. Determine the image distance, the magnification of the image, the image height, the properties of the image.

Known :

The focal length (f) = -30 cm

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

The object height (h) = 10 cm

The object distance (do) = 60 cm

Wanted : The image distance (di), the magnification of image (m), the image height (hi) and the properties of image

Solution :

Formation of image by diverging lens : The image distance (di) :

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

di = -60/3 = -20 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 = -(-20)/60 = 20/60 = 1/3

The image 1/3 smaller than the object.

Th image height (hi) :

m = hi / ho

hi = m ho = (1/3)10 = 10/3 = 3.3 cm

Properties of image

– virtual because the rays do not pass through the image

– upright

– the image smaller than the object

– the image distance smaller than the object distance

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