Lens formula and Lens maker's formula

 

LENS FORMULA:

Observe the above figure,

Let OO^I be the object, F₁ and F₂ are two foci on either sides of the lens and P is the optic centre.

Now, we are considering two rays to show the formation of image.

Ray1 is parallel to principal axis and after refraction it is passing through focus (F₁). Ray2 is passing through optic centre so it is un deviated.

Let the two rays will meet at I^I. When I^I is extended normally on to the principal axis let it meets at I. So, image formed is II^I.

Let PO, PI and PF₁ be object distance, image distance and focal length respectively.

Now let us observe the geometry of the above figure.

triangle PP^IF₁ and triangle F₁II^I are similar triangles,

 

PP^I/II^I = PF₁/F₁I .................(1) (ratio of similar sides will be equal)

But from the figure

F₁I = PI – PF₁

substituting F1I in equation (1) above, we get

PP^I/II^I = PF₁/(PI – PF₁) .................(2)

We have another set of similar triangles OO^IP and PII^I.

From these triangles we get, OO^I/II^I = PO/PI

but from figure, OO^I = PP^I, hence

we have

PP^I/II^I = PO/PI .................(3)

From (2) and (3), we get

PO/PI = PF₁/(PI-PF₁)

PI/PO = (PI-PF₁)/PF₁

PI/PO = PI/PF₁ – 1

On dividing the equation by PI, we get

1/PO = 1/PF₁ -1/PI

1/PO + 1/PI = 1/PF₁ .................(4)

The above equation is derived for a particular case of the object while

using a convex lens. To convert this into a general equation, we need to

use the sign convention.

According to the sign convention

PO = -u ; PI = v ; PF₁ = f

Substituting these values in equation 4, we get

1/v – 1/u = 1/f

Therefore the obtained lens formula is

1/f=1/v-1/u

LENS MAKER’S FORMULA

Imagine a point object ‘O’ placed on the principal axis of the thin lens as shown in figure. Let this lens be placed in a medium of refractive index n_a and let refractive index of lens medium be n_b.

Consider a ray, from ‘O’ which is incident on the convex surface of the

lens with radius of curvature R₁ at A as shown in figure.

The incident ray refracts at A.

Let us assume that, it forms image at Q, if there were no concave surface.

From the figure, Object distance PO = –u;

Image distance v = PQ = x

Radius of curvature R = R₁

n₁ = n_a and n₂ = n_b

Substitute the above values in the equation, n₂ / v - n₁ / u = (n₂-n₁) / R

ð  n_b / x + n_a / u = (n_b-n_a) / R₁ ...................(1)

But the ray that has refracted at A suffers another refraction at B on the

concave surface with radius of curvature (R₂). At B the ray is refracted and

reaches I on the principal axis.

The image Q of the object due to the convex surface is taken as object

for the concave surface. So, we can say that I is the image of Q for concave

surface. See figure.

Object distance u = PQ = + x

Image distance PI = v

Radius of curvature R = –R₂

For refraction, the concave surface of the lens is considered as

medium-1 and surrounding medium is considered as medium2. Hence the

suffixes of refractive indices interchange. Then we get,

n₁ = n_b and n₂ = n_a

Substituting the above values in equation n₂ / v - n₁ / u = (n₂-n₁) / R

n_a / v - n_b / x = (n_a-n_b) / (-R₂) ...................(2)

By adding (1) and (2) we get,

ð  n_a / v + n_a / u = (n_b-n_a)( 1 / R₁ + 1 / R₂)

Dividing both sides by na, We get

ð 1 / v + 1 / u = (n_b/ n_a – 1)(1 / R₁ + 1 / R₂)

We know n_b / n_a = n_ba called refractive index of lens with respect to

surrounding medium.

1 / v +1 / u = (n_ba– 1)(1 / R₁ + 1 / R₂)

This is derived for specific case for the convex lens so we need to

generalize this relation. For this we use sign convention. Applying sign

convention to this specific case we get,

1/v - 1/u = (n_ba– 1)(1/R₁ - 1/R₂).

We know that

1/v -1/u = 1/f

So, we get

1/f = (n_ba– 1)(1/R₁ - 1/R₂) ...................(3)

If the surrounding medium is air, then the relative refractive index

could be absolute refractive index of the lens.[since n_a=1]

1/f = (n– 1)(1/R₁ - 1/R₂) ...................(4)

This can be used only when the lens is kept in air.

Where n is absolute refractive index and this equation is called lens maker’s formula.

CHANGE IN FOCAL LENGTH OF A LENS WITH RESPECT TO SURROUNDING MEDIUM

 

Here the question is “What happens to the focal length of a lens when its surrounding medium is changed?”.

Let us find the answer by taking an example.

We can find answer for this by using Lens Maker’s Formula for any type of lens. Now for example let us apply it for a convex lens which is kept in water (surrounding medium).

Let n₁ be absolute refractive index of material of glass and

n₂ be absolute refractive index of water.

The refractive index of thin lens in air is given by

(1/f_a)=(n₁-1)(1/R₁–1/R₂).................(1)

Where  R₁ and R₂ are radii of curvature of two surfaces of the convex lens and f_a is focal length of lens in air.

Now, when lens is kept in water, we have to use refractive index of material of lens relative to water. This is n₁/n₂. And let focal length of lens in water be f_w

So, replacing in the above expression n₁ by n₁/n₂, we get

(1/f_w)=[(n₁/n₂)-1](1/R₁–1/R₂)………………(2)

Comparison of Eq.(1) with Eq.(2) shows that f_w>f_a.

Focal length increases.

And on substituting the values of refractive index of air and water in the above equations, we will get approximately f_w=4f_a  that is, focal length of a convex lens increases to four times when it is immersed in water.

Similarly by using the above process we can find the change in focal length of any lens when it is kept in any medium.

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