Question #16566

A diverging lens is to be used to produce a virtual image one-third as tall as the object. Where the object should be placed

Expert's answer

All virtual images formed by diverging lenses are upright images. Thus,& M > 0 and the magnification gives

M = -p/q = 1/3,

or

q = -p/3

Then, from the thin lens equation,

1/p - 3/p = -2/p = 1/f,

or

p = -2f = 2|f|.

So, the object should be placed at distance& 2|f| in front of the lens.

M = -p/q = 1/3,

or

q = -p/3

Then, from the thin lens equation,

1/p - 3/p = -2/p = 1/f,

or

p = -2f = 2|f|.

So, the object should be placed at distance& 2|f| in front of the lens.

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