Power of a Lens

The power of a lens tells how strongly a lens converges or diverges light rays.
A lens with smaller focal length bends light more strongly and therefore has greater power, while a lens with larger focal length bends light less and has smaller power.
This concept is important in spectacles, cameras, microscopes, telescopes, and numerical problems based on ray optics.

The deviation of the incident light rays produced by a lens on refraction through it, is a measure of its power.

or

The power of a lens is defined as the reciprocal of its focal length. It is represented by the letter P.

OR

The power (P) of a thin lens is equal to the reciprocal of its focal length (f) measured in metres.

The SI unit of power of a lens is the dioptre.
One dioptre is the power of a lens whose focal length is 1 metre.

1D = 1m⁻¹

Power of lens (in D) = \[\frac{1}{\text{focal length (in metre)}}\]

or

P = \[\frac {1}{f}\]

or

P = \[\frac {1}{f (m)}\]

Power of a Lens in a Medium:

P = (n₂ - n₁)\[\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\] = \[\frac {n_1}{f}\]

Interpretation

• A positive power indicates a converging lens.
• A negative power indicates a diverging lens.
• A greater numerical value of power means a stronger bending of light.

1. When a lens is placed in a liquid, light refracts at both curved surfaces.
2. The total power of the lens is the sum of the powers of the two surfaces: P = P₁ + P₂
3. Power of the first surface: P₁ = \[\frac {n_2−n_1}{R_1}\]
4. Power of the second surface:
   P₂ = \[\frac {n_1−n_2}{R_2}\]
5. Adding them,
   P = \[\frac{n_2-n_1}{R_1}+\frac{n_1-n_2}{R_2}\]
6. Taking (n₂ − n₁) common,
   P = \[(n_2-n_1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\]
7. Since power is related to focal length by
   P = \[\frac {n_1}{f}\]

This is the power of a lens when immersed in a medium of refractive index n₁.

Result

• If n₂ > n₁​: Lens keeps its nature, but its power decreases.
• If n₂ = n₁: P = 0, so the lens has no focusing power (becomes invisible).
• If n₂ < n₁: the lens changes its nature (convex behaves like concave and vice versa).

Note:

• Convex lens: Positive power (f > 0)
• Concave lens: Negative power (f < 0)

(i) Find the power of the lens

Given: Focal length, f = 0.5 m

Formula: P = \[\frac {1}{f}\]

Substitute: P = \[\frac {1}{0.5}\] = +2 D

Answer: +2 dioptre

(ii) Find the refractive index of glass

Given:

• f = +12 cm
• R₁ = +10 cm
• R₂ = −15 cm

Lens maker's formula:

\[\frac{1}{f}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\]

Substitute the values:

\[\frac{1}{12}=(n-1)\left(\frac{1}{10}+\frac{1}{15}\right)\]

Solving gives:

n = 1.5

(iii) Find the focal length in water

Given:

• Focal length in air = 20 cm
• n_glass = 1.5
• n_water = 1.33

For air:

\[\frac{1}{20}=0.5\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\]

For water:

\[\frac{1.33}{f}=(1.5-1.33)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\]

Using the first equation in the second, we get:

f = 78.2 cm

Answer: The focal length increases in water because the refractive index difference between glass and water is smaller.

• Spectacles use lenses of suitable power to correct vision defects.
• Cameras use lens systems to focus clear images.
• Microscopes and telescopes use lenses with suitable focal lengths and powers for magnification.