• Assume 'small aperture' $\Rightarrow$ Paraxial is valid.

• i.e. $M$ is close to $P$

• $\Rightarrow \alpha, \beta, \gamma, i$ and r are small

• $\Rightarrow \tan \alpha \approx \sin \alpha \approx \alpha$.

• $\tan \beta \approx \sin \beta \approx \beta$, etc.

• Refracted light can ba minimized by using Anti-reflection coatings.

• Angles

• $i=\alpha+\beta$

• $r=\beta-\gamma$

• $\alpha \approx \tan \alpha=\frac{M D}{O D} \approx \frac{M D}{O P}$

• $\beta \approx \tan \beta-\frac{M D}{C D} \approx \frac{M D}{C P}$

• $\gamma \approx \tan \gamma=\frac{M D}{I D} \approx \frac{M P}{I P}$

• $\therefore i = \frac{MD}{OP} + \frac{MD}{CP}$-----(3)

• $r = \frac{MD}{CP} \frac{MD}{IP}$-----(4)

• Using Snell's Law

• $\frac{\sin i}{ \sin r} = \frac{n_2}{n_1} \Rightarrow n_1 \sin i = n_2 \sin r$

• For small i,r : $\sin i \approx i, \sin r \approx r \Rightarrow n_1 i = n_2 r$-----(5)

• $\therefore n_1 (\frac{MD}{OP} + \frac{MD}{CP} = n_2 (\frac{MD}{CP} \frac{MD}{IP})$-----(6)

• Refraction at a spherical surface :

• $n_1 (\frac{MD}{OP} + \frac{MD}{CP} = n_2 (\frac{MD}{CP} \frac{MD}{IP}) \rightarrow (eqn-6)$

• $\frac{n_1}{OP} + \frac{n_2}{IP} = \frac{(n_2-n_1)}{CP} \rightarrow (eqn-7)$

• Applying 'sign convention',

• OP = -u, CP = R, IP = v

• $\frac{n_1}{-u} + \frac{n_2}{v} = \frac{n_2 - n_1}{R}$

• $\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} \rightarrow (eqn-8)$

• Gives a relation between object distance and image distance in terms of r.i and R.O.C of the spherical surface.

• Determine the position of the image when the point object is at a distance of (i) $100 {cm}$,

• (ii) $50 {cm}$ (iii) $25 {cm}$

• in front of the spherical surface.

• Solution: $\frac{n_2}{v}-\frac{n_1}{u}=\frac{(n_2-n_1)}{R}$---(1)

• $u_1=-100 {cm}, R=25 {cm}$

• $n_1=1.0, n_2=1.5$

• $\frac{1.5}{v}-\frac{1.0}{(-100)}=\frac{1.5-1.0}{2.5}=\frac{1 }{50}$

• gives $v=150 {cm}$

• i.e Real image in glass medium.

• (2) $u=-25 {cm}$

• $\frac{1.5}{v}-\frac{1.0}{(-25)}=\frac{1}{50}$

• $\frac{1.5}{v}=\frac{1}{50}-\frac{1}{25}=-\frac{1}{50}$

• $v=-75 {cm}$, virtual image

• "Thin Lens" $OA \approx OP \approx OB$

• $IA \approx IP \approx IB$

• 't' is small

• AB = thickness of the lens.

• medium.1 $\rightarrow$ Air $n_1 = 1$

• medium.2 $\rightarrow$ Glass material of r.i $n_2$

• $\frac{n_2}{v_1} - \frac{n_1}{u} = \frac{n_2 - n_1}{R_1} \rightarrow (1)$

• $\frac{n_1}{v} - \frac{n}{u} = \frac{n_1 - n_2}{R_2} \rightarrow (2)$

• Adding eq(1) and eq(2)

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

• $\frac{1}{v}-\frac{1}{u}=(\frac{n_2}{n_1}-1)(\frac{1}{R_1}-\frac{1}{R_2}) \rightarrow (4)$ constant for a given lens

• For large object distances, $\frac{1}{u} \rightarrow 0 \therefore \frac{1}{v} = constant$.

• $\Rightarrow$ Image point is fixed, independent of $u$, and is called principal focus F. The corresponding image distance is called the focal length, $f$.

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

• From (4) & (5).

• $\frac{1}{v}-\frac{1}{u}=\frac{1}{f} \rightarrow(6)$

• Lens formula

• $\frac{1}{v}-\frac{1}{u}=\frac{1}{f} \text { (Lens formula) }$

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

• for $u \rightarrow \infty$, rays from object are almost parallel to the axis, and $v=f$, the focal length.

• For a lens immersed in a liquid

• of $r . i \cdot n_l, \frac{1}{f_l}=(\frac{n_2}{n_l}-1)(\frac{1}{R_1}-\frac{1}{R_2})$

• Since $n_l>n_{air}(=1)$,

• $f_e >f_{air}$

• For common applications of

• a lens, $n_1=n_{air}=1$ $n_2=n$, r.i. of the material of lens

• $\therefore \frac{1}{f}=(n-1)(\frac{1}{R_1}-\frac{1}{R_2}) \rightarrow$ (Lens-maker's formula.)

• The formula indicates to the choice of $R_1$ & $R_2$ to obtain a desired focal length.

• For a symmetric biconvex lens, $R_1 = -R_2 = R$

• $\Rightarrow \frac{1}{f}$ = $(n-1)(\frac{1}{R}$ - $\frac{1}{(-R)})$ = $\frac{2}{R} (n-2)$

• n>1 $\Rightarrow$ f>0 $\Rightarrow$ "converying lens"

• For a symmetric biconvex lens,

• $\frac{1}{f} =\frac{2}{R}(n-1)$

• $f =\frac{R}{2} \frac{1}{(n-1)} \rightarrow(1)$

• For a symmetric biconcave lens,

• $R_1=-R, R_2=R$ gives

• $f=-\frac{R}{2} \frac{1}{(n-1)} \rightarrow(2)$

• since n>1, f< 0 in Equation(2)

• $\Rightarrow$ Diverging lens

• for n=1.5 , f=R and for n=2, f=$\frac{R}{2}$

• Normally $n_2>n_1$. if $n_1>n_2$?

• for $n_2<n_1$

• Is |$f_2$| = |f|?

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

• $\frac{1}{f_2} = (\frac{n_2}{n_1} - 1)(\frac{1}{R_2} =- \frac{1}{R_1})$

-$\frac{1}{f}$

• $\therefore$ |$f_2$| = |f|, is long as

• $n_1$ is the same (medium) on both sides of the lens, even through $R_1 \neq R_2$.

• Thus, a lens has two principal focus.

• $F_1$ is the first principal focus, $f_1 \rightarrow$ first focal length

• $F_2$ is the second principal focus,

• $f_2 \rightarrow$ second focal length

• $F_1$ and $F_2$ are equidistant from the lens.

• Normally, when we refer to the. "focus of a lens", wa refer to the point $F_2$, and the focal length $f_2$ Then, what is the importance of $F_1$ ?

• $\triangle A B P$ and $\triangle A'B' P$ are similar.

• $\frac{A B}{B P}=\frac{A' B'}{P' B'}$ or $\frac{A' B'}{AB}=\frac{B'P}{B P}$:

• Applying sign convention :

• $-\frac{h'}{h}=\frac{v}{-u}$ or $m=\frac{h'}{h}=\frac{v}{u}$

• $\triangle A B P$ and $\triangle A'B'P$ are similar;

• $\therefore \frac{A'B'}{A B} =\frac{B'P}{B P}$

• i.e $\frac{h'}{h}=\frac{-v}{-u}=\frac{v}{u}$

• or $m = \frac{v}{u}$ as before.

• Rays 1,2 and 3 appear to come from point A

• $\therefore$ A'B' is a virtual image.