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Q27:
How are power and focal length of a lens related? You are provided with two lenses of focal length 20 cm and 40 cm respectively. Which lens will you use to obtain more convergent light?

Solution :

Step 1: Theoretical Relationship between Power and Focal Length

The power ($P$) of a lens is defined as the reciprocal of its focal length ($f$) when expressed in meters. Mathematically, this is represented by the formula:

$P = \frac{1}{f (\text{in meters})}$

[Per the definition of optical power in the International System of Units (SI)]

The SI unit of power is the Dioptre ($D$), where $1D = 1m^{-1}$. This relationship implies that power is inversely proportional to the focal length: as the focal length increases, the power decreases, and as the focal length decreases, the power increases.

Step 2: Analysis of the Given Lenses

We are provided with two convex lenses (assuming convergence) with the following specifications:

  • Lens A: $f_1 = 20\text{ cm} = 0.2\text{ m}$
  • Lens B: $f_2 = 40\text{ cm} = 0.4\text{ m}$

Calculating the power for each lens:

$P_1 = \frac{1}{0.2\text{ m}} = +5.0\text{ D}$

$P_2 = \frac{1}{0.4\text{ m}} = +2.5\text{ D}$

Step 3: Visual Representation of Convergence

Lens ($f=20\text{cm}$) Higher Convergence

Step 4: Determining the Degree of Convergence

The convergence of a lens is directly proportional to its ability to bend incoming parallel rays of light. A lens with higher power has a greater ability to deviate light rays towards the principal axis [By the Law of Refraction/Snell's Law].

Lens Focal Length ($f$) Power ($P$)
Lens A 20 cm +5.0 D
Lens B 40 cm +2.5 D

Since the power of Lens A ($5.0\text{ D}$) is greater than the power of Lens B ($2.5\text{ D}$), Lens A possesses a higher capacity to converge light rays.

Final Solution: The power of a lens is the reciprocal of its focal length ($P = 1/f$). To obtain more convergent light, you must use the lens with the shorter focal length, which is the 20 cm lens, as it has a higher optical power.


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