Refraction Formation by a Lens

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$, with

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

m = $\frac{h'}{h} = \frac{v}{u}$

Converging Lens

Image Formation

Diverging Lens

A $2 {cm}$ long needle is placed erect at a distance of $10 {cm}$ in front of a thin biconcave lens of focal length $10 {cm}$.

1. Determine the position and the size of the image

2. Draw a ray diagram (with appropriate numbers for the object distance, image distance, ste.) showing formation of the image.

$u=-10 {cm}$

$f = -10 {cm}$

$\frac{1}{v}=\frac{1}{f}+\frac{1}{u}$

$=-\frac{1}{10}-\frac{1}{10}=-\frac{2}{10}=-\frac{1}{5}$

$v=-5 {cm}$

$m=\frac{v}{u}=\frac{-5 {cm}}{-10 {cm}}=0.5$

size of the image = $0.5 \times 2 {cm}$

$=1 {cm}$

The "converging" (or "diverging") capability of a lens is quantified by the parameter "Power of a lens", $P$

Intuitively, the converging capability $\propto \frac{1}{f}$

$\therefore Power \propto \frac{1}{f}$

Defined as $P=\frac{1}{f}, f$ in meter

Unit is $m^{-1}$, also called 'Dioptre' (D)

Thus, for example, a convex lens of focal length $50 \mathrm{~cm}$ has power -

$P=\frac{1}{0.5 m}=2 D$

Similarly a concave lens of focal length $40 {cm}$ has power -

$P=\frac{1}{-0.4 {m}}=-2.5 {D}$

In common terms, the unit 'D' is dropped, and people tend to use +2,-2.5, etc.

$P=+2 \Rightarrow+2 D \Rightarrow f=50 {cm}$

$P=-1 \Rightarrow-1 D$

$\Rightarrow$ Convex lens

$\Rightarrow f=-100 {cm}$

$\Rightarrow$ Concave lens.

Consider two thin lenses $L_1$ and $L_2$ of focal lengths $f_1$ and $f_2$, placed in contact:

what would be the focal length of the combination?

Since the lenses are "thin', we assume the optical centres to coincide midway between the two lenses, at $P$.

For the image formed by $\mathrm{L}_1$

$\frac{1}{v_1}-\frac{1}{u}=\frac{1}{f_1}$

For the image formed by $\mathrm{L}_2$

$\frac{1}{v}-\frac{1}{v_1}=\frac{1}{f_2}$

$I_1$ is the virtual object for $L_2$

Adding Equations (1) and (2).

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f_1}+\frac{1}{f_2}=\frac{1}{f}$ (say)

i.e. with $\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}$, we have $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$\Rightarrow$ the same form as the thin lens formula for a lens of focal length 'f'.

Since $\frac{1}{f} = P$, power of the lens

$P = P_1+P+2$

$P_1 = \frac{1}{f_1}, P_2 = \frac{1}{f_2}$

$\Rightarrow$ The combination behaves like an equivalent lens of focal length 'f'.

$f_1, f_2, f_3$ and $f_4$ may have different signs, in. +va. or -va

$\therefore P_{eq}$ is the algebraic sum of the powers of individual lenses.

What is the focal length of the combination of two thin lenses - a convex lens of focal length $30 {cm}$, and a concave lens of focal length $20 \mathrm{~cm}$ ? Is the combination 'converging' type or diverging type?

(Exercise from Text Book)

Does it matter if we interchange the positions of $L_1$ and $L_2$ ?

1. Determine the position and size of the image.

2. Draw qualitatively the corresponding ray diagram, showing the formation of the image.

$m=\frac{h'}{h}=\frac{v}{u}=\frac{-\frac{40}{3}}{-40}$ = $\frac{1}{3}$

$\Rightarrow \text { size of the image }=\frac{1}{3} \times 1.2 {cm}$

$=0.4 {cm}$

What if the lenses are separated by a distance of $5 {cm}$ ?

$m=\frac{2}{5}, v=-14 {cm}$