How can we derive the formulae for the lens ie 1/f = 1/v + 1/u ?

Hello,

This lens formula given by 

To derive first we go through refraction through the spherical refracting surface.

• A refracting surface which forms a part of a sphere of transparent refracting material is called a spherical refracting surface.

• The above figure shows the geometry of formation of image I of an object O and the principal axis of a spherical surface with centre of curvature C and radius of curvature R.

Assumptions:

(i) The aperture of the surface is small compared to other distance involved.

(ii) NM will be taken to be nearly equal to the length of the perpendicular from the point N on the principal axis.

For ΔNOC, i is the exterior angle.

∴ i = ∠NOM + ∠NCM

Similarly, r = ∠NCM − ∠NIM

i.e.,

By Snell’s law,

n₁ sini = n₂ sinr

For small angles,

n₁i = n₂ r

Substituting the values of i and r from equations (i) and (ii), we obtain

Applying new Cartesian sign conventions,

OM = − u, MI = + v, MC = + R

Substituting these in equation (iii), we obtain

This equation holds for any curved spherical surface.

Refraction by a Lens

Figure (a)

Figure (b)

Figure (c)

The above figure shows the image formation by a convex lens.

Assumptions made in the derivation:

• The lens is thin so that distances measured from the poles of its surfaces can be taken as equal to the distances from the optical centre of the lens.

• The aperture of the lens is small.

• The object consists only of a point lying on the principle axis of the lens.

• The incident ray and refracted ray make small angles with the principle axis of the lens.

A convex lens is made up of two convex spherical refracting surfaces.

The first refracting surface forms image I of the object O [figure (b)].

Image I₁ acts as virtual object for the second surface that forms the image at I [figure (c)]. Applying the equation for spherical refracting surface to the first interface ABC, we obtain

A similar procedure applied to the second interface ADC gives

For a thin lens, BI₁ = DI₁

Adding equations (i) and (ii), we obtain

Suppose the object is at infinity i.e.,

OB → ∞ and DI → f

Equation (iii) gives

The point where image of an object placed at infinity is formed is called the focus (F) of the lens and the distance f gives its focal length. A lens has two foci, F and, on either side of it by the sign convention.

BC₁ = R₁

CD₂ = −R₂

Therefore, equation (iv) can be written as

Equation (v) is known as the lens maker’s formula.

From equations (iii) and (iv), we obtain

As B and D both are close to the optical centre of the lens,

BO = − u, DI = + v, we obtain

Equation (vii) is known as thin lens formula.