 # Lens Maker’s Formula

## What is the Lens maker’s formula?

Lens is a refracting device, consisting of a transparent material. It can have two curved surfaces or one curved and one plane surface. Generally, lenses can be classified as converging (convex) and diverging lenses (concave). Lens maker’s formula relates the focal length, radii of curvature of the curved surfaces, and the refractive index of the transparent material. The formula is used to construct lenses with desired focal lengths. The formula is applicable to both types of lenses. The sign convention should be followed in the application of the lens maker’s equation.

### Focal Length and Radius of Curvature Definition

When parallel light rays are incident on a lens, the refracted rays converge to a point (for a converging lens) or appear to diverge from a point (for a diverging lens). This point is called the focus of the lens. The distance between the optical center and the focus is called the focal length.

The curved surfaces of a lens belong to two spheres. The radii of these spheres are called the radii of curvature of the lens. Depending on the shape of the lens, the radii change.

### Lens Maker Equation

The lens maker formula for a lens of thickness d and refractive index $\mu$  is given by,

$\frac{1}{f}$ = ($\mu$ - 1) $\left [ \frac{1}{R_{1}} - \frac{1}{R_{2}} + \left ( 1 - \frac{1}{\mu} \right ) \frac{d}{R_{1} R_{2}}\right ]$

Here, $R_{1}$ and $R_{2}$ are the radii of curvature of the two surfaces.

For a thin lens, the thickness d is taken to be zero. The lens maker equation for a thin lens is given by,

$\frac{1}{f}$ = ($\mu$ - 1)  $\left ( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right )$

### General Equation of a Convex Lens

A convex lens of negligible thickness is considered in the above figure. The refractive index of the lens is $mu_{2}$   and it is kept in a medium of refractive index $mu_{1}$. A point object O is kept in front of the lens. An image I' is formed due to refraction at the first surface with radius of curvature $R_{1}$ . This intermediate image serves as an object for the second surface with radius of curvature $R_{2}$. The final image I is formed due to refraction at the second surface. The object and the images lie on the principal axis. The principal axis intersects the surfaces at points C and D. The optical center of the lens is at P. Since the lens is thin, the points C,D and P are considered to be overlapping.

According to the sign convention for refraction at a curved surface,

• Distances, measured along the direction of incident light, are positive.

• Any distance, opposite to the direction of incident light, is negative.

If an object (at a medium with refractive index $mu_{1}$) is placed at a distance u, in front of a spherical surface of refractive index $mu_{2}$, having a radius of curvature R, an image is formed at a distance v from that surface such that,

$\frac{\mu_{2}}{v}$ - $\frac{\mu_{1}}{u}$  = $\frac{\mu_{2} \mu_{1}}{R}$

For refraction at the first surface, object distance OC $\approx$ OP = - u and image distance is I'CI'P=v'. Radius of curvature is $R_{1}$. Then the object-image relation due to refraction at the first surface gives,

$\frac{\mu_{2}}{{v}'}$ - $\frac{\mu_{1}}{u}$  = $\frac{\mu_{2} \mu_{1}}{R_{1}}$

The intermediate image serves as the object for the second surface. Therefore, object distance is I'DI'P=v' and final image distance is IDIP=v. Radius of curvature is negative i.e. - $R_{2}$. Applying the object-image relation due to refraction at the second surface,

$\frac{\mu_{2}}{v}$ - $\frac{\mu_{1}}{{v}'}$ = $\frac{\mu_{2} \mu_{1}}{R_{2}}$

$\frac{1}{v}$ - $\frac{1}{u}$ = $\left ( \frac{\mu_{1}}{\mu_{2}} - 1 \right )$   $\left ( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right )$

If the object is at infinity, the image is formed at the focus i.e. u=∞ and v=f. This gives,

$\frac{1}{f}$ = $\left ( \frac{\mu_{1}}{\mu_{2}} - 1 \right )$   $\left ( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right )$

If the ambient medium is taken to be air i.e. $\mu_{1}$ $\approx$ 1  and $\mu_{2}$ = $\mu$ is considered, the lens maker formula can be given in the usual form.

### Lens Maker Formula for Concave Lens and Convex Lens

For a concave lens, $R_{1}$ is negative and $R_{2}$ is positive. The lens maker formula for concave lens is given by,

$\frac{1}{f}$ = - $\left ( \frac{\mu_{1}}{\mu_{2}} - 1 \right )$   $\left ( \frac{1}{R_{1}} + \frac{1}{R_{2}} \right )$

For a convex lens, $R_{1}$ is positive and $R_{2}$is negative. The lens maker formula takes the following form,

$\frac{1}{f}$ = - $\left ( \frac{\mu_{1}}{\mu_{2}} - 1 \right )$   $\left ( \frac{1}{R_{1}} + \frac{1}{R_{2}} \right )$

### Solved examples

1. The focal length of a lens, made up of glass, is 5 cm in air. What would be the focal length of the same lens in water? The refractive indices of glass and water are 1.51 and 1.33 respectively.

Solution: The radii of curvature of the lens are $R_{1}$ and $R_{2}$. The focal length of the lens in air is $f_{a}$ = 5 cm. The refractive index of glass is  $\mu _{g}$  = 1.51. Applying lens maker equation for air,

$\frac{1}{f_{a}}$ = ($\mu _{g}$ - 1)  $\left ( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right )$

$\frac{1}{5cm}$ = (1.51 - 1) $\left ( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right )$

$\left ( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right )$ =   $\frac{1}{2.55cm}$

If the focal length is $f_{w}$in water, the lens maker equation gives,

$\frac{1}{f_{w}}$ = $\left ( \frac{\mu_{g}}{\mu_{w}} - 1 \right )$   $\left ( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right )$

Here, $\mu_{w}$ =1.33 is the refractive index of water. Substituting the value of $\left ( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right )$,

$\frac{1}{f_{w}}$ =  $\left ( \frac{1.51}{1.33} - 1 \right )$ $\frac{1}{2.55cm}$

$f_{w}$ = 18.84cm

The focal length in water is 18.84 cm.

1. An equiconvex lens is made up of a material with refractive index 1.5 and it has a radius of curvature of 20 cm. What is the focal length?

Solution: The radii of curvature of the two surfaces are equal i.e.  $R_{1}$ = -  $R_{2}$ = R. Refractive index =1.5 and R=20 cm. If the focal length is f, applying lens maker formula,

$\frac{1}{f}$ = ($\mu$ - 1) $\frac{2}{R}$

$\frac{1}{f}$ = (1.5 - 1) $\frac{2}{20cm}$

f=20 cm

The focal length is 20 cm.

### Did you know?

• If a convex lens, having focal length f, is cut along the principal axis, both the resulting pieces will have the same focal length f

• Water droplets can be considered as convex lenses and the lens maker formula is applicable.

• The reciprocal of the focal length is called the refractive power that has units dioptre (inverse meter).