AA real object and its real inverted image are to be 40 m ap

A.)A real object and its real inverted image are to be 4.0 m apart. There are two possible locations for the lens relative to the position of the object. What is the location of the object when the image is twice the size of the object?

B.)What is the other possible location? (When the image is NOT twice the size of the object.)

C.) What is the focal length of the lens?

Solution

The distance between real object and its real inverted image u+v = 4.0 m    ---------( 1)

Image size = 2(object size)

Magnification m = image size / object size

                        = 2

We know m = v/ u

                v = mu

                  = 2u

Substitute this in equation(1) you get , u +2u = 4

       3u = 4

         u = (4/3 )m

the location of the object when the image is twice the size of the object is (4/3)

(B).Other possible location v = 4-(4/3)

                                          = (12-4) / 3

                                          = (8/3)m

(C).Focal length f = [d ² -a ²]/4d

Where d = u+v = 4 m

           a = v-u = (8/3)-(4/3) =(4/3)

Substitute values you get , f = [4 ² -(4/3) ²]/[4(4)]

                                         = 0.8888 m