physics-class-12-unit-09-chapter-09-optics-resolving-power-of-optical-instruments-l-9-9-qylbxy-qezy

### <Success style="font-size:25px"> Optics Resolving Power Of Optical Instruments L-9 </Success>
### <primary style="font-size:24px"> Airy Pattern </primary>
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- 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

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<footer style = "font-size: 18px">Optics Resolving Power of Optical Instruments $\rightarrow$ Diffraction by a Circular Aperture $\rightarrow$ <span style = "color:blue"> Airy Pattern </span> $\rightarrow$ Airy Pattern $\rightarrow$ Aperture Size </footer>

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### <primary style="font-size:24px"> Airy Pattern </primary>
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- $\theta_1$ - 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$

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<footer style = "font-size: 18px">Diffraction by a Circular Aperture $\rightarrow$ Airy Pattern $\rightarrow$ <span style = "color:blue"> Airy Pattern </span> $\rightarrow$ Aperture Size $\rightarrow$ Spot Size </footer>

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- **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 $ = $\frac{1.22 \times 500 \times 10^{-9} \times 20 \times 10^{-2}}{5 \times 10^{-2}}=2.4-4 \mu{\mathrm{m}}$

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![image](https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/physics-class-12-unit-09-chapter-09-optics-resolving-power-of-optical-instruments-l-9_9-qylbxy_qezy-150-0661.2.jpg)

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<footer style = "font-size: 18px">Aperture Size $\rightarrow$ Spot Size $\rightarrow$ <span style = "color:blue"> Diffraction </span> $\rightarrow$ Resolving Power of Optical Instruments $\rightarrow$ Telescope </footer>

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### <primary style="font-size:24px"> Resolving Power of Optical Instruments </primary>
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- 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?

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<footer style = "font-size: 18px">Spot Size $\rightarrow$ Diffraction $\rightarrow$ <span style = "color:blue"> Resolving Power of Optical Instruments </span> $\rightarrow$ Telescope $\rightarrow$ Compound Microscope </footer>

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- **Angular Magnification**

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

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<footer style = "font-size: 18px">Diffraction $\rightarrow$ Resolving Power of Optical Instruments $\rightarrow$ <span style = "color:blue"> Telescope </span> $\rightarrow$ Compound Microscope $\rightarrow$ Human Eye </footer>

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- $ L - \text { Tube Length } $

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

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

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<footer style = "font-size: 18px">Resolving Power of Optical Instruments $\rightarrow$ Telescope $\rightarrow$ <span style = "color:blue"> Compound Microscope </span> $\rightarrow$ Human Eye $\rightarrow$ Resolution </footer>

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### <primary style="font-size:24px"> Limit of Resolution </primary>
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- **"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."

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<footer style = "font-size: 18px">Human Eye $\rightarrow$ Resolution $\rightarrow$ <span style = "color:blue"> Limit of Resolution </span> $\rightarrow$ Limit of Resolution $\rightarrow$ Resolution of Human Eye </footer>

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### <primary style="font-size:24px"> Limit of Resolution </primary>
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- $\Delta \theta_{\min }=1.22 \frac{\lambda}{D}$ (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 $s_{\text {min }}=f \Delta \theta_{\text {min }}=1.22 \frac{\lambda}{10} \cdot f$

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<footer style = "font-size: 18px">Resolution $\rightarrow$ Limit of Resolution $\rightarrow$ <span style = "color:blue"> Limit of Resolution </span> $\rightarrow$ Resolution of Human Eye $\rightarrow$ Example-2 </footer>

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### <primary style="font-size:24px"> Resolution of Human Eye </primary>
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- $\Delta \theta_{\min }=1.22 \frac{\lambda}{D} \rightarrow$ 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 $\mathcal{L}$.

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

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<footer style = "font-size: 18px">Limit of Resolution $\rightarrow$ Limit of Resolution $\rightarrow$ <span style = "color:blue"> Resolution of Human Eye </span> $\rightarrow$ Example-2 $\rightarrow$ Example-2 </footer>

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### <primary style="font-size:24px"> Example-2 </primary>
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- Consider two bright LED bulbs, separated by a distance $s^{\prime}$, glowing at a distance of $2 \mathrm{~km}$. from an observer. What if the minimum separation (between the bulbs) for which the observer will be able to distinguish them?
- Assume that the diameter of the pupil (of the observers eye) is 5 mm, and the wavelength of emission of the LED is $500=\text {nm.}$

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<footer style = "font-size: 18px">Limit of Resolution $\rightarrow$ Resolution of Human Eye $\rightarrow$ <span style = "color:blue"> Example-2 </span> $\rightarrow$ Example-2 $\rightarrow$ Resolving Power of Microscope </footer>

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### <primary style="font-size:24px"> Example-2 </primary>
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- $ \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}$

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<footer style = "font-size: 18px">Resolution of Human Eye $\rightarrow$ Example-2 $\rightarrow$ <span style = "color:blue"> Example-2 </span> $\rightarrow$ Resolving Power of Microscope $\rightarrow$ Compound Microscope </footer>

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### <primary style="font-size:24px"> Resolving Power of Microscope </primary>
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- 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.

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<footer style = "font-size: 18px">Example-2 $\rightarrow$ Example-2 $\rightarrow$ <span style = "color:blue"> Resolving Power of Microscope </span> $\rightarrow$ Compound Microscope $\rightarrow$ Limit of Resolution of a Microscope </footer>

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

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<footer style = "font-size: 18px">Resolving Power of Microscope $\rightarrow$ Compound Microscope $\rightarrow$ <span style = "color:blue"> Limit of Resolution of a Microscope </span> $\rightarrow$ Limit of Resolution of a Microscope $\rightarrow$ Example-3 </footer>

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

- 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 $\mathrm{mm}$.

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<footer style = "font-size: 18px">Compound Microscope $\rightarrow$ Limit of Resolution of a Microscope $\rightarrow$ <span style = "color:blue"> Limit of Resolution of a Microscope </span> $\rightarrow$ Example-3 $\rightarrow$ Example-3 </footer>

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### <primary style="font-size:24px"> Example-3 </primary>
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- A particular object, which is to be observed under a microscope, has a periodically corrugated surface of period $1 \mu \mathrm{m}$. The objective lens of the microscope has a diameter of $5 \mathrm{~ mm}$ and focal length $10 \mathrm{~ mm}$. If the object is observed with light of wavelength $600 \mathrm{~ nm}$, will the corrugations be seen clearly? (yes or no? Justify your answer).

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<footer style = "font-size: 18px">Limit of Resolution of a Microscope $\rightarrow$ Limit of Resolution of a Microscope $\rightarrow$ <span style = "color:blue"> Example-3 </span> $\rightarrow$ Example-3 $\rightarrow$ Example-3 </footer>

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### <primary style="font-size:24px"> Example-3 </primary>
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- $ \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}$

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<footer style = "font-size: 18px">Limit of Resolution of a Microscope $\rightarrow$ Example-3 $\rightarrow$ <span style = "color:blue"> Example-3 </span> $\rightarrow$ Example-3 $\rightarrow$ Thank You </footer>