Resolving Power of a Diffraction Grating

Transcription

1 Resolving Power of a Diffraction Grating When measuring wavelengths, it is important to distinguish slightly different s. The ability of a grating to resolve the difference in wavelengths is given by the ratio called the Chromatic Resolving Power R, R 1/R is the relative error We will now derive an expression for the Resolving Power of a diffraction grating. For two nearly equal wavelengths, the angular location of their maxima (except the central max) will be shifted slightly.

2 Resolving Power of a Diffraction Grating We know how a small change in will lead to a small change in the location of the maxs,. We can calculate this by taking the differential of the following relation, dsin m Taking the differential and consider small but finite changes, m dcos m or d cos A reasonable criterion for resolvable peaks from two slightly different wavelengths is to have the peak of one to sit at least at or farther away from the 1 st minimum of the other. D grating_2_lasters

3 Resolving Power of a Diffraction Grating From our phasor discussion in relation to the multiple minima between adjacent maxima in a multi-slit diffraction pattern, the phase difference between adjacent minima is given by 2/N, where N is the number of slits. 2 N 1 st adjacent min

4 Resolving Power of a Diffraction Grating We also know how phase difference is related to the angular location, dsin 2dsin or 2 Consider small changes, we have the following relationship, 2 d cos For a small change in phase 2 N, the angular position will change by, 2 2 d cos N Ndcos This gives the ½ width of the peak at order m.

5 Resolving Power of a Diffraction Grating Nd cos 1 st adjacent min

6 Resolving Power of a Diffraction Grating So, equating this width with the shift due to, Nd cos d m cos Canceling the common factor dcos, we finally arrived at the desired relation, R Nm So, the resolving power of a diffraction grating increases by using a grading with a larger number N of lines (or rulings) measuring the spectra line at a higher order m

7 Resolving Power (again) for Circular Apertures image will not be sharp Because of diffraction, light spreads out after passing thru circular apertures this imposes resolution limits to commonly used optical instruments, such as microscopes and telescopes.

8 Resolving Power for Circular Apertures Consider two non-coherent point sources (so that they don t interfere), i.e. two distant stars, angular separation between the two stars star 1 star 2 lens from telescope We will observe two diffraction patterns on top of each others.

9 Resolving Power for Circular Apertures The overlap of the two diffraction pattern might prevent one from discerning the two sources of light. A workable criterion is called the Rayleigh s Criterion which is similar in spirit to our discussion for the resolving power for the diffraction grating: The two diffraction pattern can be resolvable if the central max from one pattern is at least as far as the 1 st min of the other image. For circular aperture with diameter D, the location of the its 1 st order diffraction minimum is: sin 1.22 ( 1.22 is a geometric factor) D 1

10 Resolving Power for Circular Apertures The Limit of Resolution for a circular aperture is defined as the smallest angular separation between two light sources that can be resolved according to the Rayleigh s Criterion and it is given by: sin min 1.22 D star 1 min star 2 lens from telescope

11 Resolving Power for Circular Apertures A optical device such as a telescope or microscope will have a high Resolving Power if its has a small Limit of Resolution so that nearby objects with a small angular separation can be resolved. This gives the following ways to increase the Resolving Power: increase the diameter D use a bigger len/mirror in telescope decrease the wavelength use a shorter wavelength of light in chip production

12 Example 36.6: Resolving Power of a Camera Lens Given: f=50 mm f-number of f/2 object distance 9.0m wavelength = 500nm What is the minimum distance between two points on the object that one can resolve? f-number = f/d D = f/f-number = 50mm/2 =25 mm Rayleigh s Criterion gives: m sin D 2510 m min min 3 5 rad

13 Example 36.6: Resolving Power of a Camera Lens For a simple lens, we know that the angular separation of two points on the object is given by, s y y s y s y' s ' y separation of object points y separation of the corresponding image points s object distance s image distance

14 Example 36.6: Resolving Power of a Camera Lens Applying the minimum condition, we have, y rad y m rad mm s (2.410 ) 0.22 (on object) On the camera film, the image separation will be approximately, y ' rad y ' 50 mm(2.410 rad) s ' mm (on film) s ~ f if image is focused on the film

15 X-Ray Diffraction X-Rays were discovered by Wilhelm Rontgen in There were indication that x-rays are EM waves with a very short wavelength, ~ m. Max Von Laue in 1912 proposed that scattered x-rays from the crystalline solid might produce a diffraction pattern when they interfere with each others. The successful experiment verify that: 1. x-rays are waves 2. crystalline solids are arranged in regular repeating patterns

16 X-Ray Diffraction 2D simplification Conds for constructive interference: a = r 2dsin m, m1,2, (Bragg cond.)

17 X-Ray Diffraction X-Ray Diffraction has been very useful in exploring the crystalline structure of solids and structures of organic molecules Rosalind Franklin was the first to use x-ray diffraction to image a DNA molecule in The dark crossed bands was the first evident of the helical structure of the DNA molecule.