1. KIN2  .30 deuterium t min =-0.123 GeV 2 q’=2.78 GeV P’ d =.352 Gev Inner calo 10x9 block R=13.5 cm   ’ =8° t min =-0.33 GeV 2 q’=2.72 GeV P’ d =.579 Gev The Problem How to detect the maximum of recoil deutons with reasonable silicon detector Without interfere with the accepted experiment PYB

2. Reasonnable silicona+/- 40° angular range ===> close of the Target as possible 1)Move the beam  it is possible let said 1cm 2)Detector close to 1cm from the target 2 cm Silicon detector dimension +/- 2xtan(40°)/sin(18)=+/-5.42cm which is limit of reasonable 20° This insure that all the dvcs events produced in the target and with -t min <t<0.33GeV 2 And with   270° are emitted in front of the silicon. We favor the Beth-Heither process PYB

3. DD DD T D MEV -t GeV 2 D Q=k-k’ 18 o  PYB

4. LD2 SILICON 1 Stop in LD2  E=0 2 Stop in Si  E>> MeV 3 Stop after Si  E<10 MeV T Mev  E=0 1 2 3 At angle  fixed cos   E=0 1 2 3 At T fixed  Note: With a Silicon thickness of 1mm (  E) 2 ~5-10MeV. In the silicon of Compton Polarimeter (  E) 3 ~0.3 MeV PYB

5. Conclusion: it is possible to do a  full  distribution between -80 o to+80 o f for the bin [2.6,3.] GeV 2 in –t. in this bin  E is >3MeV  10 times bigger that in the Compton Polarimeter PYB

6. 2 cm 16x(5.5x50 mm2) strips 32 strips Silicon strips: 1mm 100 mm PYB

7. ARS Charge amplifier. + derivation Using ARS allow: Times coincidence ~ 1-2 ns  E analyze Vacuum feed-through Like this of the polarimeter Compton Only the charge amplifier require a new study PYB

8. Proton from the electro-desintegration of the deuton e+D  e+n+p Are the two dominant backgrounds Moller electrons The proton field was computed by Pavel :

9. PYB

10. 15 cm LH2 I=10  A 32 stripes : 5,5 x100 mm 2 at 2 cm from the beam PYB But some stop in the deuterium  T p < 21 MeV

11. 2 cm Tan(  )=d/l=2/16   =7.12° Electron Backgroung : The forward part of the silicon must have difficulties. If it the case 1,2 or n strips can be forgotten The best will be to use a shorter target 5 or 4 cm   = 18 o PYB But that will implied to increase beam intensity to 20 µA and this intensity require the use of the raster !!!!!!

12. Possible Improvement 20 o Target  mm

13. An other setup 18 o X 0 D=8cm

15. But an any case the absorption in the LH2 will conduct at the same range in t and 

16. My feeling: 1)It is technically doable* for KIN2 2)It will be better to used a 4 or 5 cm target !! 3)The range in t and  is limited. In this range a cross section can be extract In this range the B-H is dominant “ Le jeu en vaut il la chandelle ?” Knowing that this technique will be and more difficult to be applied at the 12 GeV when Q 2 and t will increase. mechanical support of the silicon vacuum feed –through position of target versus beam Preamplifier * PYB