Question

For a converging lens, an object placed at half of the focal length will produce an image which is:

Real and half the size of the object

Virtual and twice the size of the object

Real and twice the size of the object

Virtual and half the size of the object

None of these scenarios

Answer 1

Answer:

The image will be Virtual and twice the size of the object

Because if the object distance value is less than one focal length produces image that are upright and virtual. And the image size is double the value of object or equals to the focal length.

I/f = 1/u + 1/v which implies that v = uf/u-f . Here the object size u = f/2 then

Image size v = [ (f/2) . f] / [f/2 - f ] = -f . The negative indicates the image is virtual and f indicates the image size is equal to the one focal length of twice the size of the object (i.e., v = 2u = 2 (f/2) = f ).