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The POWER of a lens (in diopters, D) is defined as
A$P = f$ (focal length in metres)
B$P = 1/f$ where $f$ is in centimetres
C$P = f^2$
D$P = 1/f$ where $f$ is in metres
Answer & Solution
Correct answer: D. $P = 1/f$ where $f$ is in metres
1. NCERT §9.6.3 defines lens power.
2. $P = 1/f$, where $f$ is the focal length expressed in METRES.
3. SI unit: diopter (D), defined as $1\,\text{D} = 1\,\text{m}^{-1}$.
4. Higher power = shorter focal length = stronger converging (or diverging) action.
5. Sign: positive $P$ for converging lens (converging combination), negative $P$ for diverging.
6. Example: a lens with $f = +50\,\text{cm} = +0.5\,\text{m}$ has $P = +2\,\text{D}$. A diverging lens with $f = -25\,\text{cm}$ has $P = -4\,\text{D}$.
7. Options A and C have wrong forms. Option D uses cm instead of m, which gives a number 100× too large.
_Source: NCERT Class 12 Physics Part 2, Ch 9, §9.6.3 (Power of a Lens, Eq. 9.21), p. 18._
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