Question

Derive the Lens Maker’s formula.

Answer 1

Hint: A lens is composed of two identical spherical surfaces having the same refractive index different from the surrounding medium. We will have to first apply the formula for refraction at the interface to find out the image distance formed by the first interface. Then that image will act as a virtual object for the second surface. Then on applying the refraction formula at the second surface, and equating both the equations, we will get the Lens Maker formula.

Formula used:

Image formed by refraction of a light ray at the interface of two mediums with different refractive indices,  ${\displaystyle \frac{n_2}{v}}-{\displaystyle \frac{n_1}{u}}={\displaystyle \frac{n_2-n_1}{R}}$ .

Complete answer:

Let us assume a double convex lens made by two joining two identical spherical surfaces with radius of curvature R. Where the refractive index of the surrounding medium is  $n_1$  and that of the glasses be  $n_2$ . Image distance be v and the object distance from the lens be u.

Therefore, on applying the formula for refraction of light on a spherical surface, we will get

 ${\displaystyle \frac{n_2}{v}}-{\displaystyle \frac{n_1}{u}}={\displaystyle \frac{n_2-n_1}{R}}$ , where OB = -u, is the object distance, OI = +v is the image distance and BC = +R is the radius of curvature of the glass.

The formation of the final image by a convex lens happens in two steps. The refraction first takes place at the interface ABC, then the 1st image will form at $I_1$  which will then serve as a virtual image for the second interface ADC.

Now, applying the above equation for surface ABC, we get

 ${\displaystyle \frac{n_1}{OB}}+{\displaystyle \frac{n_2}{B{I}_1}}={\displaystyle \frac{n_2-n_1}{B{C}_2}}$  ………. (i)

Similarly, we can write the equation for second interface as

 $-{\displaystyle \frac{n_2}{D{I}_1}}+{\displaystyle \frac{n_1}{DI}}={\displaystyle \frac{n_2-n_1}{D{C}_2}}$ 

Here, the lens is thin, so  $B{I}_1=D{I}_1$ . Upon adding the equations (i) and (ii), we get  ${\displaystyle \frac{n_1}{OB}}+{\displaystyle \frac{n_1}{DI}}=(n_2-n_1)({\displaystyle \frac{1}{B{C}_1}}+{\displaystyle \frac{1}{D{C}_2}})$ .

Here, the object has been assumed to be at infinity, so OB will be very large

 $\therefore {\displaystyle \frac{n_1}{DI}}=(n_2-n_1)({\displaystyle \frac{1}{B{C}_1}}+{\displaystyle \frac{1}{D{C}_2}}$ 

Dividing both sides by  $n_1$ , we get

 ${\displaystyle \frac{1}{DI}}=({\displaystyle \frac{n_2}{n_1}}-1)({\displaystyle \frac{1}{B{C}_1}}+{\displaystyle \frac{1}{D{C}_2}})$ 

As,  $D{C}_2=-R_2,DI=f$  and  $B{C}_1=R$ 

Then we get  ${\displaystyle \frac{1}{f}}=({\displaystyle \frac{n_2}{n_1}}-1)({\displaystyle \frac{1}{R_1}}-{\displaystyle \frac{1}{R_2}})$ .

Hence, we get the Lens Maker’s formula.

Note: Sign convention for the distances is very important and should be the same while application of formula on both the interfaces. Drawing separate diagrams and separately applying the formula on both surfaces will reduce the chances of confusion a lot.