(i) The lens X is a concave (diverging) lens (negative power indicates concave lens).
Focal length: $f = \frac{1}{P} = \frac{1}{-2.5} = -0.4 \text{ m} = \mathbf{-40 \text{ cm}}$
This lens is prescribed for myopia (short-sightedness).
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(ii) Given: $m = -2$, object distance $u = -20$ cm (by sign convention).
Since $m = \frac{v}{u}$: $-2 = \frac{v}{-20}$, so $v = +40$ cm.
(I) Nature: Real and inverted (m is negative).
(II) Size: Image is twice the size of the object (|m| = 2, enlarged).
(III) Position: Image is formed 40 cm on the opposite side of the lens from the object.
(IV) Sign of image height: Negative (inverted image has negative height by sign convention).
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(iii) Lens A (focal length 10 cm) shows a higher degree of convergence/divergence.
Reason: Power $P = \frac{1}{f}$; smaller focal length means greater power, hence greater ability to converge or diverge light.
Source: Chapter 9, Sections 9.3.4, 9.3.6
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