Effective Focal Length

Reflection and Refraction of light produces some beautiful scenery around us

The formation of Rainbow is due to the Refraction of Light

The reflection of the setting sun on the water produces a stunning visual treat

Humans have also artificially created some of the life-altering things based on these phenomena

The simple plane-mirror in which we see our reflection almost everyday, is based on the Reflection of Light

Spectacles, helping people with a defect in their eyes to see, uses Refraction of Light

Let us see the Image Formation Process in a Lens

If we consider a convex lens, we know the ray of light coming from infinity will merge at the focus after refraction

We know that the rays of light passing through the optical centre of the lens remains undeviated

And the ray of light passing through the focus will become parallel to the principle axis after refraction

Combining these two rays, the image formed will be real and inverted in nature

The image distance will be given by the lens law $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ Wherev= Image Distance u= Object Distance f= Focal Legth of the Lens

Let us now move forward and see the Image Formation by combining two lens

Let us join two lenses $L_{1}$ and $L_{2}$ with focal lengths, $f_{1}$ and $f_{2}$

Had $L_{2}$ been not there, let the image by $L_{1}$ be at $I_{1}$ , such that the image distance is $v^{'}$

Using the lens formula for $L_{1}$ , $\frac{1}{f _{1}} = \frac{1}{v ^{'}} - \frac{1}{u}$

For the lens $L_{2}$ , the virtual object will be at $I_{1}$ and the image formed will be at $I$

So, by using the same lens formula for lens $L_{2}$ , $\frac{1}{f _{2}} = \frac{1}{v} - \frac{1}{v ^{'}}$ we will get the focal length of the lens $L_{2}$

Now we replace the combination by a single lens and we need the focal length of the lens that will place the image at the same point $I$

By using the lens formula for lens $L$ we get the focal length for L, $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$

Adding equation (1) and (2), $\frac{1}{f _{1}} + \frac{1}{f _{2}} = \frac{1}{v ^{'}} - \frac{1}{u} + \frac{1}{v} - \frac{1}{v ^{'}}$ Thus, $\frac{1}{f _{1}} + \frac{1}{f _{2}} = \frac{1}{v} - \frac{1}{u}$

Therefore, for the combination of two lenses the effective focal length is given by, $\frac{1}{f _{1}} + \frac{1}{f _{2}} = \frac{1}{f}$

For combination of n lenses, $\frac{1}{f _{1}} + \frac{1}{f _{2}} + . . . . . . = \frac{1}{f}$

Revision

Effective focal length for combination of two lenses is, $\frac{1}{f _{1}} + \frac{1}{f _{2}} = \frac{1}{f}$

Effective focal length for n lenses will be given by, $\frac{1}{f _{1}} + \frac{1}{f _{2}} + . . . . . . = \frac{1}{f}$

The End