The intensity distribution due to Fraunhofer diffraction by a circular aperture is given by

$$

I=I_0\left[\frac{2 J_1(v)}{v}\right]^2, v=\frac{\pi}{\lambda} 2 a \sin \theta

$$

$J_1(\vartheta)$ is Bessel function of the first order.

$\sim$ 86% of the energy is contained within the Airy Disk

$\theta_1$ - is the angle corresponding to the first minima

$\Rightarrow v_1=\frac{\pi}{2} 2 a \sin \theta_1=3 \cdot 832$

$\Rightarrow \sin \theta_1=\frac{3 \cdot 832}{2 \pi a} \lambda=1 \cdot 22 \frac{\lambda}{2 a}$

$\therefore \tan \theta_2 \approx \theta_1 \Rightarrow \frac{w}{f}=1 \cdot 22 \frac{\lambda}{2 a}$

$\text { given 2w} =1.22 \frac{\lambda f}{a}$

$1.22 \frac{\lambda}{2 a} \ll 1$

$\therefore \sin \theta_1 \approx \theta_1 \approx \tan \theta_1$

$\mathrm {W} = 1.22 \frac{\lambda}{2a} \mathrm { f }$

$\mathrm { W } \sim 100 \mu \mathrm { m }$

$r_o=1.22 \frac{\lambda}{D} \cdot f$

$r_o \sim 1 \mu m$

$r_o \rarr\text { Spot size }$

1. Diffraction by Circular Aperture

$\lambda=500 \mathrm{~ nm}, f =20 \mathrm{~ cm}, 2 a=1 \mathrm{~ mm}$

$\mathrm {W} =1.22 \frac{\lambda}{2 a}$

$f =\frac{1.22 \times 500 \times 10^{-9} \times 20 \times 10^{-2}}{1 \times 10^{-3}}$ = $122 ~ \mu m ~ (0.122 \mathrm{~mm})$

2. Diffraction by the Lens (Aperture)

$\mathrm {r} =1.22 \frac{\lambda}{D} \cdot f$ = $\frac{1.22 \times 500 \times 10^{-9} \times 20 \times 10^{-2}}{5 \times 10^{-2}}=2.4-4 \mu{\mathrm{m}}$

In Geometrical Optics (or Ray Optics), we discussed 3 Optical Instruments:

Telescope

Microscope

Human Eye

$\rightarrow$ Details of image formation and Magnification (or Magnifying Power) $\rightarrow$ See Lectures 8 & 9

$\rightarrow$ Now Resolving Power $\rightarrow$ Why now?

Angular Magnification

$m=\frac{\beta}{\alpha}=\frac{\left(h^{\prime} / f_2\right)}{\left(-h^{\prime} / f_o\right)}=-\frac{f_o}{f_e}$

$L - \text { Tube Length }$

$f_o - \text { Focal length of Objective }$

$f_e - \text { Focal length of Eyepeice }$

$r_o = f \times \left(1.22 \frac{\lambda}{D}\right)$

$\rarr$ finite size, even for a point source S.

"Rayleigh Criterion"

Linear separation between the intensity peaks.

$\Delta \theta_{\min }=1.22 \frac{\lambda}{D}$ is the minimum resolvable angular separation

$S_{\text {min }}=f \Delta \theta_{\text {min }}$ is the minimum linear separation in the focal plane of the objective lens

$\frac{1}{s_{\min }} \rightarrow$ "Resolving Power."

$\Delta \theta_{\min }=1.22 \frac{\lambda}{D}$ (rad.)

1. 'Limit of resolution' refers to the minimum angular separation $\Delta \theta_{\min }$ for which two objects are "just resolved" (i. e., distinguishable) on the focal plane of the objective (lens).

2. According to Rayleigh Criterion, two objects are said to be just resolved when the central diffraction maximum of one object coincides with the first diffraction minima of the other object.

3. The linear separation on the focal plane of the objective lens $s_{\text {min }}=f \Delta \theta_{\text {min }}=1.22 \frac{\lambda}{10} \cdot f$

$\Delta \theta_{\min }=1.22 \frac{\lambda}{D} \rightarrow$ Minimum angular separation

$s_{\min} = f \Delta\theta_{\min}$ for two objects to be resolved

$s^{\prime}=L \Delta \theta \rightarrow$ Linear separation at a large

$S=f \Delta \theta \quad$ distance $\mathcal{L}$.

$D - \text { Diameter of the pupil} (2-8 \mathrm{~ mm})$

$\Delta \theta_{\text {min }} =1.22 \frac{\lambda}{D}$

$=\frac{1.22 \times 500 \times 10^{-9}}{5 \times 10^{-3}}$

$=1.22 \times 10^{-4} \mathrm{rad}$

$s_{\text {min }}^{\prime}= L \Delta \theta_{\text {min }}=1$

$= 2 \times 10^3 \times 1.22 \times 10^{-4} \mathrm{rad}$

$= 2.44 \times 10^{-1} \mathrm{~m}$

$=24.4 \mathrm{~cm}$

Refers to the ability of a microscope to resolve (or distinguish) the smallest possible features of an object.

If $s_{\text {min}}$ is the minimum separation that can be measured between two "just resolved" features of an object or a sample, then

$s_{\text {min}}$ is called the limit of resolution of, and

$\frac {1} {s_{\text {min}}}$ is called the resolving power of the microscope.

$\sin \Delta \theta_{\min }=1.22 \frac{\lambda}{D}$ for $P$ and $Q$ to be "just resoloved"

$u=f_o$

$\frac{\lambda}{D} \ll 1 \rightarrow \sin \Delta \theta_{\min } \approx \Delta \theta_{\min }$

$\therefore s_{\min }=f_o \Delta \theta_{\min }=1 \cdot 22 \frac{\lambda}{D} \cdot f_o$

$f_o -$ focal length of the objective.

$s_{\text {min }}=1.22 \frac{\lambda}{D} f_o$

1. The expression for $s_{\text {min }}$ looks identical to that obtained for the telescope. In the case of a telescope, the emphasis was on increasing D to get a better resolution In the case of a microscopes, one chooses an objective of shorter focal length to get a better resolution. (Why?)

2. Choice of a shorter wavelength to observe the object would also improve the resolution.

Normally, $f_o$ few $\mathrm{mm} \rightarrow$ few $\mathrm{cm} ; D \sim$ few $\mathrm{mm}$.

$\Delta \theta_{\min } =1.22 \frac{\lambda}{D}$

$$s_{\min }= f_o \Delta \theta_{\mathrm{min}}=\frac{10 \times 10^{-3} \times 1.22}{5 \times 10^{-3}} \times \lambda$$

$2 \times 1.22 \times 600 \times 10^{-9} \mathrm{~ m}$

$12 \times 1.22 \times 10^{-7} \mathrm{~ m}$

$1.464 \times 10^{-6} \mathrm{~ m}=1.464 \mu \mathrm{m}$

$s_{\min} = 1.464 \mu ~ m$

No, cannot be resolved