The lens is a transparent medium bounded by two surfaces and at least one of them must be curved. A lens is said to be thin if the gap between the two surfaces is very small. A lens will be converging with positive focal length, and diverging if the focal length is negative. Therefore from this, we can conclude that a convex lens need not necessarily be a converging and a concave lens diverging. Every lens has some specific value that we can compute by using the lens makers formula. This topic will explain it with examples. Let us learn it!

**Lens Makers Formula**

**What is the lens makers formula?**

Real lenses have the finite thickness between their two surfaces of curvature. An ideal thin lens with two surfaces of equal curvature will have zero optical power. It means it will neither converge nor diverge light. A lens with some thickness which is not negligible is called a thick lens.

Lenses are of two types based on the curvature of the two optical surfaces. Convex and concave.

Lens maker’s formula is the relation between the focal length of a lens to the refractive index of its material and the radii of curvature of its two surfaces. It is used by lens manufacturers to make the lenses of particular power from the glass of a given refractive index.

The lens is thin, therefore the distance measured from the poles of the two surfaces of the lens can be taken to be equal to the distances measured from the optical center.

**Formula**

Lens maker formula is used to construct a lens with the specified focal length. A lens has two curved surfaces, but these are not exactly the same. If we know the refractive index and the radius of the curvature of both the surface, then we can determine the focal length of the lens by using the given lens maker’s formula:

\(\frac{1}{f} = (\mu -1) \times (\frac{1}{R_1} – \frac{1}{R_2})\)

Where,

f | The focal length of the lens |

\(\mu\) | Refractive index |

\(R_1 and R_2\) | the radius of the curvature of both surfaces |

It is noteworthy that the lens should be thin so that the separation between the two refracting surfaces should be small. Also, the medium on either side of the lens should be the same.

**Solved Examples**

Q.1: Find out the focal length of the lens whose refractive index is 2. Also, the radius of curvatures of each surface is 20 cm and -35 cm respectively.

Solution:

Given parameters are:

\(\mu = 2\),

\(R_1\) = 20 cm

and \(R_2\) = – 35cm

Lens maker’s formula is:

\(\frac{1}{f} = (\mu -1) \times (\frac{1}{R_1} – \frac{1}{R_2})\)

\(\frac{1}{f} = (2-1) \times (\frac{1}{20} – \frac{1}{-35})\)

\(\frac {1}{f} = 1 \times (0.05 + 0.028)\)

\(\frac {1}{f}\) = 0.078

Therefore, f = \(\frac {1} { 0.078}\)

f = 12.82 cm

Q.2: Value of the refractive index of lens is 2.5. The curved surfaces are having the radius of curvatures 10 cm and -12 cm respectively. Find out the focal length of the lens.

Solution:

Given parameters are,

\(\mu\) = 2.5,

\(R_1\) = 10 cm

and \(R_2\) = -12 cm

Lens maker’s formula is:

\(\frac{1}{f} = (\mu -1) \times (\frac{1}{R_1} – \frac{1}{R_2}) \)

\(\frac{1}{f} = (2.5 -1) \times (\frac{1}{10} – \frac{1}{-12}) \)

\(\frac {1}{f}\) = 0.274

So, f = 3.64 cm

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