IMGS Solution Set #9

Transcription

1 IMGS Solution Set #9 1. A white-light source is filtered with a passband of λ 10nmcentered about λ nm. Determine the coherence length of the light emerging from the filter. Solution: The coherence length is determined from the frequency bandwidth ν ν c c ν λ λ λ coherence length c ν λ (600 nm) 36 μm mm λ 10 nm. Light transmitted through an object composed of two slits separated by 00 μm is observed on an observation screen at a distance of z 1m. The observed pattern consists of eight interference fringes spread over the distance x 0mm.Determine the wavelength λ 0 of the source. Solution: D 0 d 0 λ 0 z λ 0 D 0d 0 z 0 mm D 0.5mm 8 λ 0.5mm 00 μm 1m 500 nm 3. Use Brewster s law to find the refractive index of a glass for which light from the green Hg line (λ nm)reflected at an angle of θ is linearly polarized. Solution: θ B radians tan 1 n If n 1 1(air), then n n n 1 tan[0.31 0π] n 1 1

2 4. Linearly polarized light with intensity I 0 (proportional to the time average of the squared magnitude) and oriented parallel to the angle θ relative to the x-axis passes through an ideal linear polarizer oriented along the x-axis. The light emerging from the first polarizer passes through a second polarizer oriented along the direction of the original polarization (i.e., at angle θ relative to the x-axis). Show that the output intensity is I I 0 4 [θ] Solution: The Jones vector for the incident linearly polarized light is: E in E 0 θ sin θ and the intensity is: I 0 E 0 θ +sin [θ] E 0 The matrix for the first polarizer is M θ [θ] sin[θ][θ] sin [θ][θ] sin [θ] The matrix for the second polarizer is: M 0 [0] sin [0] [0] sin [0] [0] sin [0] The output is: E out [θ] sin[θ][θ] sin [θ][θ] sin [θ] E 3 θ 0 E θ sin θ 0 θ θ sin θ E 0 θ sin θ and the output intensity is: could also just use Malus law twice: I out E 0 θ θ sin θ θ sin θ I in 4 θ 1 I out I in 4 [θ] I out I in [θ] I 1 I 0 [θ] I I 1 [θ] I I I1 [θ] [θ] 4 [θ] I 0 I 1 I 0 I I 0 4 [θ]

3 5. Circularly polarized light of intensity I 0 is incident on a sandwich of three ideal linear polarizers. The first and third are crossed and the angle of the second polarizer is θ relative to the axis of the first. Show that the output intensity is I 1 [θ] sin [θ] Two solutions given: (1) The Jones vector for circularly polarized light is: E E 0 1 ±i so the intensity is: E E (E ) T E E 0 1 i 1 ±i E 0 I 0 I ll use the matrix operator for linear polarizer at angle θ M 0 [0] sin [0] [0] sin [0] [0] sin [0] M θ [θ] sin[θ][θ] sin [θ][θ] sin [θ] M π π sin π π sin π π sin π Theoutputvectoris: E out M π M θ M 0 E 0 ±i 0 0 E 0 [θ] 0 1 sin [θ][θ] sin[θ][θ] sin [θ] ±i E 0 0 θ sin θ The output intensity is: I out E 0 0 θ sin θ 0 θ sin θ I 0 θ sin θ µ 1 I out I in [θ]sin [θ] E 0 [θ]sin [θ] () Circularly polarized light is changed into linearly polarized light by the first polarizer, so the intensity is reduced by half. I 1 1 I 0 [θ] I h π i I 1 θ ³ I I 0 sin [θ] h π i [θ]+sin ³ h π i θ I 1 I 0 I I 1 1 [θ] sin [θ] I I 0 1 [θ] sin [θ] h π i sin [θ] 3

4 6. ewton s rings are observed when the convex surface of a lens is placed upon an optically flat surface and illuminated with monochromatic light. A pattern of circular fringes is observed. the radius of curvature of the convex surface is R 4mand the semidiameter of the first observed bright ring is r 0 1mm. Determine the wavelength of the source. The phase difference of light reflecting from the flat surface (rare-to-dense) and from the curved surface (dense-to-rare) has a phase difference of: φ s + π x R π + π λ 0 The result is a bright fringe if φ is an even integer multiple of π radians and dark if φ is an odd integer multiple of π radians. In the problem statement, the distance from the center to the first bright ring is 1mm 4π λ 0 (1 mm) 4000 mm + π π 4 λ 0 λ 0 (1 mm) 4000 mm 1 (1 mm) 4000 mm mm 1 μm 500 nm 4

5 7. A Michelson interferometer is set up to observe straight fringes and is illuminated with λ 0 64 nm. (a) As the path length in one arm is lengthened, 00 fringes are seen to cross the center of the field of view. Determine the distance that the mirror moved. zzzzzzz This one is too easy... if the mirror is moved one-half wavelength, then the optical path changes by twice that or one full wavelength, which results in one fringe passing the center of the field of view. Since 00, z 00 λ0 100λ nm 6.4 μm mm (b) If a monochromatic light source with λ nm is substituted and the mirror is moved the same distance as in part (a), determine the number of fringes that will pass the center mm λ mm λ mm 454 nm 74.9 fringes 5

6 8. A lens with focal length f 10 mm is used to image a narrow slit of width d 0 00 μm using light from a sodium source with λ 589.3nm. The distance from the object to the lens is z mm and the light is observed on a screen at a distance of z 1mfrom the lens. The lens is then cut into two equal pieces along a diameter and separated by a distance of x 400 μm. Determine what is observed at the observation screen (HIT: this is a variety of interferometer). Solution: The split lens creates two images of the point source that are mutually coherent. We can find the distance z from the lens to the images of the point source from the imaging equation: f 10 mm z mm z z 1 f 300 mm 10 mm z 1 f 300 mm 10 mm 00 mm M T z z 1 3 so the separation of the images of the point source is 3 of the separation of the halves of the lens: d μm μm the distance from the images to the observation plane is: z 1m z 1m 00 mm 800 mm d 0 D 0 λ 0 z D 0 λ 0 z d 0 D mm 589.3nm 800 mm μm 1.77 mm 6

7 9. (OPTIOAL BOUS PROBLEM) A sandwich of +1 ideal lossless polarizers is arranged so that the difference in angle of the axis of on polarizer and that of the adjacent polarizer is α. Inthisway, the angle of the last polarizer is θ α relative to that of the first polarizer. Linearly polarized light with intensity I 0 is incident upon the first polarizer with its polarization at the same angle as the first polarizer. For the case where is very large, find a good approximation for the output intensity as a function of α. (hint: Taylor series for ine followed by binomial series). Consider explicitly the case where θ α π and. Solution: Recall Taylor Series: Ã X! d n f f [x] dx n (x x 0) n n! n0 xx0 f [x 0] + df (x x 0) 1 + d f 0! dx 1! dx xx0 (x x 0) + + dm f! dx xx0 m (x x 0) m + m! xx0 and the binomial expansion: (1 x) n 1 nx + n (n 1) x! + +( 1) r n! (n r)!r! xr + We can expand the ine function using the Taylor series: α µ1 α! + α4 4! ow apply Malus law: n (n 1) (n ) x 3 3! I 1 I 0 α I I 1 α I 0 α α I 0 4 α I I 0 α I0 ( α) Substitute the Taylor series for the ine into Malus law: ( α) µ1 α! + α4 4! µ µ α 1! α4 4! + If α is small, i.e., α ' 0, then α À α À α 4 À µ ( α) 1 α! ow use the binomial expansion where x α! and n : I I 0 ( α) µ1 α! µ µ α ( 1) α higher-order terms!!! µ α 1 because α 4 α! I 1 α 1 α 7 (α) 1 1 θ

8 ow consider the specific caseof θ α π, which means that the last polarizer is orthogonal to the first polarizer. If these were the OLY two polarizers, then the output should be zero, but the presence of the intermediate polarizers changes this significantly. Ã π! µ I I0 1 I 0 µ1 π I Obviously, this means that has to be somewhat large, since the output intensity is impossibly negative if 1or. If is smallish (say 5)and θ π,then µ I 5 I0 1 1 I 0 1 If grows towards (so that the angle increment α 0), then µ lim [I ] lim I I 0 so all of the intensity gets through if there are many ideal polarizers with a small angle between adjacent pairs. I[] Graph of the approximate output irradiance as a function of the number of polarizers for θ π radians 90 (i.e., the first and last polarizers are orthogonal). The approximation of the intensity gives a physically useful (nonnegative) result only for ' 5 and approaches 1 for large (and therefore for small α). 8