Resolving power and rayleigh criterion of resolution, Other Engineering

Assignment Help:

Q. Explain Rayleigh criterion of resolution. What is meant by resolving power of grating?

Ans. 

Resolving Power

(i) Meaning of Resolution : The resolving power of an optical instrument is tis ability to separate far away objects that are close together into individual images, as opposed to a single merged image. If we look at two stars in the sky, separated by a large enough angle one can say very easily that they are two stars by looking at them through a telescope. The telescope resolves these two stars into separate objects while our eye cannot resolve them and they look like one star to us, this is all because of diffraction. Similarly if we look at the headlights of a car as it back away from us to a far distance when the car is close, it is easy to distinguish two separate headlights. But as it gets farther away, it's harder to resolve the two headlights. Finally, there is a certain point when the car gets even farther away, that we cannot distinguish the two headlights clearly. This inability to resolve two closely spaced objects is due to diffraction. Thus to see two close objects just as separate is the resolution. In microscopes and telescopes the geometrical positions of two nearby point objects is due to diffraction thus to positions of two nearby point objects is resolved, which is known as geometrical resolution. Whereas in case of diffraction grating we resolve two nearby spectral lines which are very close to each other this is known as spectral resolution.

Resolving Power : The resolving power of an instrument is the ability of an optical instrument to distinguish (or resolve) two closely spaced objects.

(a)  Resolving power of the Telescope : The resolving power of telescope is defined as the minimum angle between two objects of equal magnitude (intensity) to see them just as separate. i.e. the smallest angular separation that can be unambiguously distinguished is called the resolving power of the telescope and is  proportional to the ration of the wavelength of light being observed to the diameter of the telescope. Thus, the larger the diameter, the smaller the minimum angular separaton and the higher the resolving power.

(b)  Resolving Power fo a Microscope : The resolving power of a microscope is defined as the smallest separation at which two separate objects can be distinguished. The resolving power of a microscope is a function of the wavelength of light used and the numerical aperture (NA) which is characteristic of the lens system and the medium between the lens and the specimen.

(c)  Rayleigh Criterion of Resolution : In order to express the resolving power of an optical instrument Lord Rayleigh proposed a universal criterion know as Rayleigh's Criterion of resolution. This criterion is the generally accepted criterion for the minimum resolvable detail.

According to Rayleigh two nearby point source images are said to be just resolved if the position of central maxima of one source coincides with the minimum of another point source. It is possible to resolve the two objects as long as the central maxima of the two diffraction patterns do not overlap. If the two central maxima overlap the two objects look like one. Let P and Q be the principle maxima of diffraction in wavelength of spectral lines of wavelength say l and l + d l. If the difference in the angle of diffraction is quite large then two maxima can be seen distinctly and both the spectral lines will appear well resolved. As shown in Fig.(a)

Now if the difference in wavelength of spectral lines decreases both the spectral lines gets closer. A situation comes when the positions of principle maxima of spectral line of wavelength l coincides with the position of spectral line of wavelength l+dl or vice versa. Then the resultant intensity of both diffraction patterns will only be observed. This is known as condition of just resolution. The resultant intensity curve is shown in Fig. (b). A dip in the intensity curve in the middle of central maxima of P and central maxima of Q is found, where the intensity is approximately 20% less than that of any intensity peak individually. This position of spectral lines where the intensity in the middle is 80% approximately of any spectral lines is known as condition of just resolution for spectral lines, also known as spectral resolution.

Now if the difference in wavelength dl is more decreases then the diffraction pattern of both the spectral lines collapse to each other and resultant intensity curves show that there is only one maxima. Hence in result we are not able to resolve these spectral lines. As shown in Fig.(c).


Related Discussions:- Resolving power and rayleigh criterion of resolution

Series of cash flows in economical engineering, Series of Cash Flows M...

Series of Cash Flows Most engineering economic analysis involve more than a single return occurring after the investment is made. In such cases, the present worth or the futur

sweep effects, In the previous discussion, we investigated the static aero...

In the previous discussion, we investigated the static aeroelastic behaviour of straight (unswept) wings. These wings are characterised by an effective decoupling of wing bending a

Mine Surveying.., Question 1 (100 marks) As Surveyor in charge of an underg...

Question 1 (100 marks) As Surveyor in charge of an underground mine you have been instructed to design a tunnel to join a crosscut to a haulage drive. The details of the crosscut a

Manual override relays, Manual override relays: When the manual-overrid...

Manual override relays: When the manual-override relays are not energised, this current flows through brushes riding on slip rings mounted on the propeller spinner bulkhead and

Bio physiology Matlab Simulink Models, Review the lecture on BB document “P...

Review the lecture on BB document “Pharmacology …..” slide 6 where the solution for a 1st order system was demonstrated. The generic system is A(X)?(K1)?B(Y)?(K1’)?C(Z) 1. Cr

Digital signal processing, compute IDFT using DIT and DFT X(K)={20,-5.828-j...

compute IDFT using DIT and DFT X(K)={20,-5.828-j2.414,0,-0.172-j0.412,0,-0.172+j0.414,0,-5.828+j2.414}

Write Your Message!

Captcha
Free Assignment Quote

Assured A++ Grade

Get guaranteed satisfaction & time on delivery in every assignment order you paid with us! We ensure premium quality solution document along with free turntin report!

All rights reserved! Copyrights ©2019-2020 ExpertsMind IT Educational Pvt Ltd