Answer: (b) The image is virtual, erect and diminished, formed between the optical centre and the focus on the same side as the object; (c) power = –4 D.
(a) Diagram: image formed by a concave lens (object beyond F1)
concave lens ////
\|/ /////
| /////
| /////
A | /////
^-----------------------------------------|//
| \\\\\\\ ::::: |
| \\\\\\\ ::::: |
| \\\\\\A':::: |
| :::::^\\\\\\ |
| ::::: ! \\\\\\\ |
---------------------------------------------|--------------------------
B F1 B' |O \\\\\\\
| \\\\\\\
| \\\\\\\
/|\ \\
Ray 1 (from A, parallel to axis) diverges after refraction as if coming from F1 (dotted : line).
Ray 2 (from A, through optical centre O) goes undeviated (\\\ line).
AB = object; A'B' = virtual, erect, diminished image between F1 and O.Construction: (1) a ray from A parallel to the principal axis, after refraction, diverges as if coming from the focus F1 (shown dotted backwards); (2) a ray from A towards the optical centre O passes undeviated. The refracted rays do not meet; when produced backwards they appear to meet at A′. Drawing A′B′ perpendicular to the axis gives the image A′B′ between F1 and O on the same side as the object.
(b) Nature of image: virtual, erect and diminished; it is formed between the optical centre O and the focus F1, on the same side of the lens as the object (for any position of the object).
(c) For a concave lens the focal length is negative: \(f = -25\ \text{cm} = -0.25\ \text{m}\).
\[P = \frac{1}{f\,(\text{in m})} = \frac{1}{-0.25} = -4\ \text{D}\]
The power of the lens is –4 dioptre (negative, as a concave lens is diverging).