Solve a problem. An object 5 cm high stands in front of a converging lens at a distance of 20 cm and an image 2 cm high is formed. Determine the optical power and focal length of the lens.

Explanation:

Height of object =5cm

Position of object, u=−25cm

Focal length of the lens, f=10cm

Position of image, v=?

We know that,

v

1

−

u

1

=

f

1

v

1

+

25

1

=

10

1

v

1

=

10

1

−

25

1

So,

v

1

=

50

(5−2)

That is,

1

=

50

3

So,

v=

3

50

=16.66cm

Thus, distance of image is 16.66cm on the opposite side of lens.

Now, magnification =

u

v

That is,

m=

−25

16.66

=−0.66

Also,

m=

heightofobject

heightofimage

or

−0.66=

5cm

heightofimage

Therefore, Height of image =3.3cm

the answer is f1 and d2 because if you put equal force on those two points the beam would balance trust me on this

answer; how far out of place which the object finished relative to the starting location;