(i) Focal Length:
From observation 3, u = −50 cm and v = +50 cm. Using lens formula:
$$\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{50} - \frac{1}{-50} = \frac{2}{50}$$
f = +25 cm
Justification (without calculation): When object is placed at 2f, image is also formed at 2f on the other side. Here |u| = |v| = 50 cm, so 2f = 50 cm, giving f = 25 cm.
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(ii) Incorrect Observation:
Observation 6 (u = −15 cm, v = +37.5 cm) is incorrect.
Since f = 25 cm and u = −15 cm, the object is placed within the focal length (between O and F). For such a position, a convex lens forms a virtual, erect, magnified image on the same side as the object, so v should be negative, not positive.
Ray Diagram: (Draw object between O and F; two rays — one parallel to principal axis refracted through F₂, one through optical centre straight — diverging rays are extended backward to meet on the same side, giving a virtual, erect, magnified image.)
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(iii) Magnification for u = −30 cm:
From table, v = +150 cm.
$$m = \frac{v}{u} = \frac{+150}{-30} = \mathbf{-5}$$
The image is real, inverted and 5 times magnified.
Source: Chapter 9, Sections 9.3.6 and 9.3.7
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