Combination of Lenses in Contact

We use binoculars to observe things that are far away.

A binocular helps us to see a distant view clearly.

Because it consists of a combination of lenses arranged which can be adjusted as per the distance.

The splitting of light when passing through a medium is known as the Dispersion of Light.

Before beginning with the power of the lens, let’s understand some basics.

A convex lens can converge a beam of parallel rays to a point on the other side of the lens.

This point is called a focal point and its distance from the centre of the lens is called the focal length.

Similarly, a concave lens diverges a straight beam of light coming from the source.

The point where the ray of light seems to refract and diverge is called principal focus and its distance from the centre of the lens is called the focal length.

Now let us learn about Power of Lens.

The power of Lens is defined as the converging or diverging capacity of the lens.

It can also be stated as the reciprocal of focal length.

Power of the lens is given by, $P = \frac{1}{f}$ Where, $f =$ focal length

S.I. unit of power is Diopters (D) or $M e t e r^{- 1} (m^{- 1})$

Power of the lens is positive for convex or converging lens while it is negative for concave or diverging lens.

Now we’ll study a combination of lens.

When more than one lens is used, the power of the lens is added.

Consider two lenses with focal lengths $f_{1}$ and $f_{2}$ are arranged as above.

The power of the system of the lens is given as, $P = P_{1} + P_{2}$ Where, $P_{1} =$ power of lens $l_{1}$ $P_{2} =$ power of lens $l_{2}$

Therefore, $P = \frac{1}{f _{1}} + \frac{1}{f _{2}}$ Where $f_{1}$ and $f_{2}$ are focal lengths of lens $l_{1}$ and $l_{2}$ respectively.

Revision

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

Therefore, $P = \frac{1}{f}$ Where $f$ is the focal length

SI unit of power of the lens is Diopter or $m e t e r^{- 1} .$

The power of the system of the lens is given as, $P = P_{1} + P_{2}$ Where, $P_{1} =$ power of lens $l_{1}$ $P_{2} =$ power of lens $l_{2}$

Therefore, $P = \frac{1}{f _{1}} + \frac{1}{f _{2}}$ Where $f 1$ and $f 2$ are focal lengths of lens $l_{1}$ and $l_{2}$ respectively.

The End