What's the image formed?

Question

An object is placed 17 cm in front of a diverging lens of focal length -11 cm. (Is there an equation used and what steps are taken to get the answer?)THANKS!

Answer 1

Rather, by drawing a ray diagram you will see tt since the object is placed BETWEEN the focal point and the 'second focal point' (which is 2 x the focal length from the verticle), the image formed will be real, magnified and inverted.

You may want to check out the other variations for objects placed BEFORE the focal point, AT the focal point, AT 2 focal points away, or BEYOND 2 focal points as well.

I dunno of any equation but i doubt there are any. the answer to ur qn is more of how does the object looks like. and to find out the size of the image it very much depends on the lens used. and perhaps you can apply some trigo.

Answer 2

the required eqn is -
(1/v)-(1/u)=(1/f)

This is called lens formula where
v stands for image distance,
u for object distance, and f for focal length.

But there is a sign convention to this, ie follow the following rules:-
a) All distances are to be measured from the optic centre of the lens.
b)Distances in the direction of incident light are positive while distances in the opposite direction are negative.
c)Heights above the principal axis are positive while those below the principal axis are negative.

So in a diverging lens, ie a concave lens, f , u and v are all negative .
On the other hand, in a converging(convex)lens, f and v are positive while u is negative.

Also magnification in a lens is given by the formula
M=v/u
Also, M=height of image/height of object
So, height of image/height of object=v/u

U can use this formula also.

But never forget the sign convention and also never forget that all this is only in the case of lenses and not mirrors.

Answer 3

Whats the size of the original object?

Answer 4

smaller, upside down, and backwards ???