Solution Set #7

Transcription

1 Solution Set #7. Consier monochromatic light of wavelength λ 0 incient on a single slit of with along the -ais an infinite length along y. The light is oserve in the Fraunhofer iffraction region at a istance L. Derive an approimate epression for the angular with at half-maimum irraiance of the central peak of the iffraction spot (i.e., the FWHM). If oserve in the Fraunhofer iffraction region, the oserve irraiance is proportional to the square magnitue of the appropriately scale Fourier transform: g [, y; λ 0,z ] z ep µ z +πi ν 0 t λ 0 F ξ = The oject function is: f [, y] =RECT h i [y] Its Fourier transform is: F [ξ,η] = SINC [ξ] [η] = [πξ] sin [η] πξ where δ [η] is a -D Dirac elta function along the vertical irection, which has finite area an infinitesimal support along the vertical irection; we may think of δ [η] as concentrating all of the energy onto the -ais. Sustitute the variales into this epression: F ξ = h i π sin = π λ 0z sin π ( ) = ³ π y y The irraiance is the square: I [, y] g [, y] SINC " # The full with at half maimum is the with of the SINC function etween the locations where its amplitue is + : SINC [u] = sin [πu] (πu) = = SINC [u] = sin [πu] πu = = The half angle θ is the value of u that satisfies this relation. We know that SINC [0] = an

2 SINC [] = 0. You can look up the appropriate value of u or solve for it y iteration: 0.5 = sin π π = π = sin [0.4π] 0.4 = = (0.4π) sin [0.45π] 0.45 = = (0.45π) sin [0.44π] 0.44 = = 0.7 (0.44π) sin [0.445π] = = (0.445π) sin [0.445π] = = (0.445π) I ll call that close enough: 0.445π =.390 raians SINC = λ0 z = λ0 z = = = λ 0z = θ = λ 0 = z = θ = θ = λ 0 This often is roune upwar so that the full with at half maimum is approimately θ = λ 0 NOTE (again) that the with of the iffraction spot varies as the reciprocal of the slit with.

3 . The Fraunhofer iffraction pattern of a pair of rectangular slits of with separate y the center-tocenter istance is illuminate y monochromatic light with λ 0 = 650 nm. The Fraunhofer iffraction pattern is viewe at the ack focal plane of a lens with focal length f = 800 mm. The center-to=center separation etween fringe maima is oserve to e D =.04 mm. The fifth maimum of the interference pattern on each sie is missing, which means that it coincies with a zero in the iffraction envelope ue to the with of the slit. Determine an. The oject function is: Ã f [, y] = RECT + RECT! + [y] The Fourier transform is: Ã Z α= + Z! α=+ + F [ξ,η] = ep [ πiαξ] α + ep [ πiαξ] α [η] α= α=+ µ µ = ep πi πiξ + µ ξ ep πi ξ [η] + µ µ ep πi + πiξ + µ ξ ep πi + ξ [η] µ µ µ µ = πiξ ep +πi ξ ep πi ξ ep +πi ξ [η] + µ µ µ µ πiξ ep πi ξ ep πi ξ ep +πi ξ [η] = ep [+iπξ]sin[πξ] [η]+ ep [ iπξ]sin[πξ] [η] πξ πξ = SINC [ξ] (ep [+iπξ]+ep[ iπξ]) [η] = SINC [ξ] cos [πξ] [η] If oserve at the Fraunhofer plane, the scale Fourier transform is: F ξ = = SINC cos π y Again, the Dirac elta function constrains the irraiance to the -ais. The -epenence of the irraiance is: # I [, y =0] 4 SINC cos "π = 4 SINC " = SINC " λ0 z # Ã! +cos π λ0 z # Ã! +cos π In our situation, the light is oserve at the ack focal plane of a lens with focal length f = 800 mm, so the The perio of the oserve cosine fringe is: D = λ 0f =.04 mm = 650 nm 800 mm = =0.5mm.04 mm The fact that the fifth maimum is missing means that the SINC function ecays to zero at that location: SINC =0if =5 = = 5 =0.mm λ0 f 3

4 3. Monochromatic light with wavelength λ 0 is incient on a circular aperture of iameter. The circularly symmetric iffraction pattern oserve at a istance L in the Fraunhofer iffraction region has raius r from the central maimum to the first zero: r =. Lλ 0 (this istance is the separation require of the iffraction patterns from two point sources for them to e istinguishe uner the Rayleigh criterion for resolution) (a) Compare this result to the linear istance from the central maimum to the first zero for a square aperture of with equal to the iameter of the circular aperture. The raius of the first zero of the iffraction pattern of the circular aperture is r =. Lλ 0 The raius to the first zero of a square aperture is the value of where: SINC Lλ0 =0 = = Lλ 0 <.Lλ 0 () If the monochromatic light at λ 0 illuminates a circular lens of iameter an focal length f, the Fraunhofer iffraction pattern is oserve at the focal plane so that L = f an : r =. fλ 0 If oserve in lue visile light, fin an approimate relation etween the iameter of the iffraction spot an the focal length of the lens. This is a very convenient rule of thum for imaging systems. r =. fλ 0 For lue light, λ 0 = 400 nm =.44λ0 = r =.44 fλ 0 =.44λ 0 = 000 nm = μm r = μm f We often efine the focal ratio (f-numer, f/#) of the lens as the ratio of the focal length to the iameter, so the iameter of the iffraction spot measure in micrometers is approimately equal to the f/# of the lens; r = f/#[μm] a smaller iameter lens leas to a larger iffraction spot. 4

5 4. Compare the iameters of the iffraction spots for telescopes with primary optics having iameters = 00 in (Hale Telescope on Palomar Mountain) an =90mm(Questar). f r Palomar = μm 000 mm = f 00 in in f r Questar = μm 90 mm =. 0 5 f the two optics ha the same focal length, the iffraction spot of the Palomar telescope woul e smaller y aout a factor of 56. This is as far as you "ha" to go, ut you can go farther. The f/numers of the two telescopes are very ifferent; the Palomar primary mirror is f/3.3 = f =3.3, while the focal length of the Questar is aout 300 mm = f/#= = 4, so the iffraction spots have raii approimately equal to 3.3 μm for Palomar an 4 μm for the Questar, which ifferyafactorofaout4insteaof56. 5