Note that this equation is valid only for thin, spherical lenses. Unless otherwi

Question

Note that this equation is valid only for thin, spherical lenses. Unless otherwise specified, a lens problem always assumes that you are using thin, spherical lenses.

The equation above allows you to calculate the locations of images and objects. Frequently, you will also be interested in the size of the image or object, particularly if you are considering a magnifying glass or microscope. The ratio of the size of an image to the size of the object is called the magnification. It is given by

All of the quantities in the above equations can take both positive and negative values. Positive distances correspond to real images or objects, while negative distances correspond to virtual images or objects. Positive heights correspond to upright images or objects, while negative heights correspond to inverted images or objects. The following table summarizes these properties:

Explanation / Answer

12 cm and 24 cm

F)
The focal length of the diverging lens ia f = -15 cm

The image has size h' = 5h/9

Where h is the height of the object

The image is upright means that the magnification is positive

the magnification m = -v/u

v is the image distance and u is the object distance

m = -v/u = h'/h
-v/u = 5/9

The equation for the lens is

1/f = 1/u + 1/v
u/f = u/v + 1
u/-15 = -9/5 + 1
u = -4*-15/5
u = 12 cm

The object is at distance 12 cm from the lens

H)

The image distance is v = -24 cm

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