(a) The lens used is a concave (diverging) lens. The image formed is virtual, erect, and enlarged on the same side as the object, which matches the behaviour of a concave lens when the object is between F₁ and O.
(b) Given: image distance v = 30 cm, magnification m = 2 (real image, so m = –2).
$$m = \frac{v}{u} \Rightarrow -2 = \frac{30}{u} \Rightarrow u = -15 \text{ cm}$$
The object was placed 15 cm from the lens (between F₁ and 2F₁).
Source: Chapter 9, Section 9.3.4
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(a) The key clue is the ray diagram — emergent rays diverge and their extensions meet on the same side as the object, forming a virtual, erect, enlarged image. This is characteristic of a concave lens (when object is between O and F₁). A convex lens gives a virtual erect image only when object is between O and F₁, but the image in that case also appears on the same side — however, the rectangular lens symbol with diverging rays in the diagram points to a concave lens.
(b) For a real image formed by a convex lens, magnification m = –v/u (negative for real inverted image). Since size of image is twice the object, |m| = 2, so m = –2. Using m = v/u: u = v/m = 30/(–2) = –15 cm. This means the object is 15 cm from the lens on the same side as the incoming light (between F₁ and 2F₁). Always use sign convention carefully in board exams.