### Lens :

Lens is a transparent medium bounded by two surfaces, at least one of them must be curved. We have to study about thin lens. A lens is said to be thin if the gap between two surfaces is very small.

### Lens maker’s formula

Consider the thin lens shown here with the two refracting surfaces having radii of curvature equal to R_{1}and R_{2} respectively.

The refractive indices of this surrounding medium and of the material of the lens are n_{1} and n_{2} respectively.

Now using the result that we obtained for refraction at single spherical surface we get,

For first surface XP_{1}Y,

$ \displaystyle \frac{n_2}{v_1} – \frac{n_1}{u} = \frac{n_2 – n_1}{R_1} $ . . . (1)

For second surface XP_{2}Y ,

$ \displaystyle \frac{n_1}{v} – \frac{n_2}{v_1} = \frac{n_1 – n_2}{R_2} $ . . . (2)

Adding (1) and (2),

$ \displaystyle \frac{n_1}{v} – \frac{n_1}{u} = (n_2 – n_1) (\frac{1}{R_1}-\frac{1}{R_2}) $

$ \displaystyle \frac{1}{v} – \frac{1}{u} = (\frac{n_2}{n_1} – 1) (\frac{1}{R_1}-\frac{1}{R_2}) $

$ \displaystyle \frac{1}{f} = (\frac{n_2}{n_1} – 1) (\frac{1}{R_1}-\frac{1}{R_2}) $

Note : A lens is converging if its focal length is positive and diverging if focal length is negative (in cartesian sign convension). From this we can conclude that a convex lens need not necessarily be a converging and a concave lens diverging

**Limitations of the lens maker’s formula:**

1. The lens should be thin so that the separation between the two refracting surface should be small.

2. The medium on either side of the lens should be same.

If any of the limitation is violated then we have to use the refraction at the curved surface formula for both the surfaces.

### Power of Lens

$ \displaystyle P = \frac{1}{f} $

where f should be in meters with proper sign i.e. +ve for convex and − ve for concave. P is in Dioptre .

$ \displaystyle \frac{1}{f} = P = (\frac{n_2}{n_1} – 1) (\frac{1}{R_1}-\frac{1}{R_2}) $

Illustration : A thin equiconvex lens of glass of refractive index n = 3/2 and of focal length 0.3 m in air is sealed into an opening at one end of a tank filled with water (n = 4/3). On the opposite side of the lens, a mirror is placed inside the tank on the tank wall perpendicular to the lens axis, as shown in figure.

The separation between the lens and the mirror is 0.8 m. A small object is placed outside the tank in front of the lens at a distance of 0.9 m from the lens along its axis. Find the position (relative to the lens) of the image of the object formed by the system.

Solution : As one side of the lens is air and another is water. We should go as per the refraction at the curved surface formula.Focal length in air

= 30 cm ⇒ |R| = 30 cm

For refraction from the first curved surface

$ \displaystyle \frac{n_2}{v} – \frac{n_1}{u} = \frac{n_2 – n_1}{R} $

On putting n_{2} = 3/2 , n_{1} = 1 , u = −90 and R = 30 , we get

⇒ v = 270 cm.

This image will behave as an object for the second surface so

$ \displaystyle \frac{4/3}{v’} – \frac{3/2}{270} = \frac{4/3 – 3/2}{-30} $

On solving

⇒v’ = +120 cm

This will behave as a virtual object for the plane mirror and will form a real image in front of the mirror at a distance of 40 cm from the mirror. This image will behave as an object for surface two of the lens.

$ \displaystyle \frac{3/2}{v”} – \frac{4/3}{-40} = \frac{3/2 – 4/3}{30} $

⇒ v” = −54cm.

This will again behave as an object, for surface one of the lens

$ \displaystyle \frac{1}{v_f}-\frac{3/2}{-54} = \frac{1- 3/2}{-30} $

⇒ v_{f} = − 90 cm

Final image will be 90 cm right of the lens.