Power of a lens

Lens formula in terms of power

How to find the power of lens using the focal length

Power of lens and its unit

Optical power

Diopter formula

Power of lens calculation

The power of a lens is specified as P = \[\frac{1}{F}\], where f is the focal length.

The S.I. unit of power of a lens is \[m^{-1}\]. This is also known as diopter.

The focal length (f) of a converging lens is considered positive and that of a diverging lens is considered negative. Thus, the power of a converging lens is positive and that of the diverging lens is negative.

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Fig.1 shows two lenses \[L_{1}\] and \[L_{2}\] placed in contact. The focal lengths of the lenses are \[f_{1}\] and \[f_{2}\] respectively. Let P be the point where the optical centers of the lenses coincide (lenses being thin).

Suppose, a point object is placed at a point O beyond the focus of lens \[L_{1}\] such that OP = u (object distance) on the common principal axis (coaxially). The first lens \[L_{1}\] alone forms an image at \[I_{1}\] where P1 = \[v_{1}\] (image distance). This point I1 works as the virtual object for the second lens \[L_{2}\] and the final image is formed at I, at a distance PI = v. The ray diagram(Fig.1) formed by the combination of two convex lenses has the following attributes:

u = Object-distance for the first lens

v = final image-distance for the second lens

\[v_{1}\] = image-distance for the first image I1 for the first lens. As the lenses are pretended to be thin, \[v_{1}\] is also the object-distance for the second lens.

The lens formula for the image \[I_{1}\] formed by lens \[L_{1}\] will be

\[\frac{1}{v_{1}}\] - \[\frac{1}{u}\] = \[\frac{1}{F}\] ….(1)

The equation for the image formation for the second lens \[L_{2}\]

\[\frac{1}{v}\] - \[\frac{1}{v_{1}}\] = \[\frac{1}{f_{2}}\]….(2)

Adding eq(1) and (2):

\[\frac{1}{v_{1}}\] -\[\frac{1}{u}\] + \[\frac{1}{v}\] - \[\frac{1}{v_{1}}\] = \[\frac{1}{F_{1}}\] + \[\frac{1}{f_{2}}\]

\[\frac{1}{v}\] - \[\frac{1}{u}\] = \[\frac{1}{F_{1}}\] + \[\frac{1}{f_{2}}\]….(3)

If the combination is replaced by a single lens of focal length F such that it forms the image of O at the position I,

\[\frac{1}{v}\] - \[\frac{1}{u}\] = \[\frac{1}{F}\]……(4)

This type of lens is called the equivalent lens for the combination.

Combining (3) and (4),

\[\frac{1}{F}\] = \[\frac{1}{f_{1}}\] + \[\frac{1}{f_{2}}\]……(5)

Here, F is the focal length of the equivalent lens for the combination. As the power of a lens is P = 1/F, eq (5) immediately gives

The power of any number of lenses in contact is equal to the algebraic sum of the power of two individual lenses.

This is true for any situation involving two thin lenses in contact.

The power of a lens is measured as the reciprocal of the focal length of the lens.

Relation with focal length: A lens of less focal length produces more converging or diverging and is said to have more power.

i.e.,

Since, P = \[\frac{1}{F}\]

We get,

Here, v = refractive index of the material

\[R_{1}\] = Radius of curvature of the first surface of the lens

\[R_{2}\] = Radius of curvature of the second surface of the lens

For a converging lens, power is taken as positive and for a diverging lens, power is taken as negative.

The S.I. unit of power is dioptre (D).

When f = 1 meter, P = \[\frac{1}{F}\] = 1/ 1 = 1 dioptre

Hence, one dioptre is the power of a lens of focal length one meter.

When f is in 1 cm, P = \[\frac{1}{F}\] / 100 = 100/ f

So we get the formulas to describe the relationship between P and f,

Optical power is defined as the degree to which a lens, mirror, or other optical system converges or diverges the light.

Optical power is also referred to as the dioptric power, convergence power, refractive power, or refractive power. It is equal to P = \[\frac{1}{F}\].

Dioptre formula is used to calculate the optical power of a lens or curved mirror. The dioptre is the unit for a measure of the refractive index of a lens. The power of a lens is specified as the inverse of the focal length in meters, or D =\[\frac{1}{F}\], where D is the power in dioptres.

Example. Find the power of a plano-convex lens, when the radius of a curved surface is 15 cm and v =1.5.

Solution: Given \[R_{1}\] = ∞, \[R_{2}\] = - 15 cm, v= 1.5 cm

P = \[\frac{1}{f}\] = (v-1) (\[\frac{1}{R_{1}}\] - \[\frac{1}{R_{2}}\])

= (1.5-1) (1/ ∞ + 1/ 15)

On solving we get,

FAQ (Frequently Asked Questions)

Q1: Which Lens has a More Power Thick or Thin?

The power of a lens depends upon the curvature and focal length (f).

The thick convex lens has more power than a thin one because the thick one has a greater curvature and less focal length.

Q2: A Person is Viewing an Extended Object. If a Converging (Convex) Lens is Placed in Front of his Eyes, will he Feel that the Size has Increased?

When the converging (convex) lens is placed in the front of his eyes, it is used as a magnifying lens and make the images large and wide. Also, the converging lens makes the images virtual and erect.

Thus the person will feel the size of the extended object increase.