(a) Draw a ray diagram to show the formation of image by a concave lens when an object is placed in front of it.

(b) In the above diagram mark the object-distance (u) and the image-distance (v) with their proper signs (+ve or -ve as per the new Cartes. ian sign convention) and state how these distances are related to the focal length (f) of the concave lens in this case.

(c) Find the nature and power of a lens which forms a real and inverted image of magnification -1at a distance of 40 cm from its optical center.

(a)

When an object is placed any-where between infinite and optical center of a concave lens, the image formed is

(i) Between O and F.

(ii) Virtual, erect.

(iii) Diminished.

(b) CB = - u

CF = - f

CB' = - v

The relation between u, v and f is given by the lens formula:

As both u and v are negative the above equation will change to

We know that the focal length of a concave lens is negative, so the above equation will be changed to,

(c) Magnification m = -1

v = +40 cm (real and inverted)

Nature of the lens = ?

Power of the lens, P = ?

According to the lens formula,

Thus, focal length = 20 cm

Because the focal length is positive, thus it is convex lens.

Since power of lens is positive, lens will be converging in nature.