Lens Makers formula

The lens is a transparent medium that is bordered by two surfaces, at least one of which must be curved in order for the lens to function. When the distance between the two sides of a lens is exceedingly short, it is said to be thin. When a lens has a positive focal length, it will be converging; when the focal length is negative, it will be diverging.

Lenses are classified into two categories based on the curvature of the two optical surfaces that make up the lens. Convex and concave shapes are used.

The lens maker’s formula is the relationship between the focal length of a lens and the refractive index of its material, as well as the radii of curvature of the two surfaces on which it is constructed. Glass with a specific refractive index is used by lens manufacturers to create lenses of specific powers using glass with a specific refractive index.

Because the lens is narrow, the distances measured between the poles of the two surfaces of the lens can be assumed to be equal to the distances measured between the optical centre and the poles of the two surfaces of the lens.

Lens Maker’s Formula Derivation 

The final image of the thing is created at the point I. The lens is surrounded by a liquid with a refractive index of 𝜇1 (aqueous solution) (in this case, air). Specifically, the lens is constructed of a material with a refractive index of 𝜇2. Let us trace two light beams starting from the object O, denoted by the letters OG and OP, in order to determine where the image I of the object O is produced. The beam OP is parallel to the primary axis of the lens (i.e., perpendicular to the spherical surface) and 𝜇2, as a result, passes through the lens without deviating from it. It is refracted at the border of the lens-air system and deflected from its initial course as a result of this deflection. Point O1 will be obtained by following the refracted beam backward until it meets the lens’s primary axis, which is point O2 (say at a distance v1 from point P). This is the position in which the object would appear if someone were to look at it via the lens from the inside. In this case, the new object position is taken to be this point, and the light ray O1H starting from this point is drawn until it reaches the point H. We may also consider the entire left portion of the system to be a medium with a refractive index of 𝜇2 at this point because we have modified the position of the object to be as if one were looking through the lens from within it. Following that, ray O1H passes through the lens and out into the medium outside (in this case, air), where it has a refractive index of 𝜇1.

because we have shifted the object’s position to make it appear as if one were looking through the lens from within The ray O1H now travels through the lens and into the medium outside (in this case, air), which has a refractive index of 𝜇1. The refractive index of the lens material is higher than the refractive index of air (i.e. 𝜇1 > 1), hence the deviation observed will be in the direction opposite to the normal drawn at the point of refraction. As a result, the refracted ray HI will be noticed at the location depicted. Once this occurs, the rays coming from point O (OI travelling through the primary axis, along with the OG + HI) collide at point I on the opposite side of the lens, where the image is produced.

𝜇2/𝑣−𝜇1/𝑢=(𝜇2−𝜇1)/𝑅

Considering the genesis to be at point P (since the lens is thin, we can take the origin at point P instead of points D and E individually for each refraction).

It should be noted that the object is located at O, the image is located at O1, and the centre of curvature for this refraction is located at C2. As a result, we have:

𝜇2/𝑣1−𝜇1/𝑢=(𝜇2−𝜇1)/𝑅1

For the second refraction at the other border of the lens, the object is now assumed to be at O1 (as previously indicated), which is at a distance v1 from the point P in the previous equation. This refraction occurs at the point I, which is on the other side of the lens and separated by a distance v. The centre of curvature for this refraction is located at C1. This results in: 

𝜇1/𝑣−𝜇2/1=(𝜇2−𝜇1)/𝑅2

Given that we are attempting to establish a formula for the focal length of the lens, let us assume that the object has now been transported to infinity (i.e., u = ∞), which we already know results in the production of an image at v = f, which is the focal point

     Therefore: 

1/𝑓=(𝜇2−𝜇1) (1/𝑅1−1/𝑅2)

CONCLUSION

Real lenses have a finite thickness between their two surfaces of curvature since they are made of two surfaces of curvature. The optical power of an ideal thin lens with two surfaces of equal curvature will be zero for a perfect thin lens with two surfaces of equal curvature.Unlike idealised thin lenses, real lenses have a finite thickness between their two surfaces of curvature. An ideal thin lens with two surfaces of equal curvature would have zero optical power, meaning that it would neither converge nor diverge light.

Also from the formula,

 optical power= 1/𝑓=(𝜇2−𝜇1) (1/𝑅1−1/𝑅2)

We can see that putting radius of curvatures(R1 and R2)=∞,

We will get optical power=   1/𝑓=(𝜇2−𝜇1) (1/𝑅1−1/𝑅2)=0

Light will not be converged or diverged as a result of this property. A thick lens is a lens that has a significant amount of thickness.