Question

1. What percentage of the light (assuming uniform and symmetric emission) is collected from a point source in air using a microscope objective with a NA = 0.45? NA = 1? The NA are given relative to air.

2. In the calculation for part a, what would your values be in water?

3. You are setting up an absorption measurement in your laboratory and need to select the optics you will use. The molecule of interest absorbs at 455 nm, and you have a 455 nm LED to use as your source. Your sample exists in a standard 1 cm quartz cuvette. What are the properties of the focusing and collection lenses (diameter, focal length, material, coatings)? Justify your answers.

Answer 1

solution : (a) The value provided in the question are: NA = 0.45 nit (ains The formula to calculate the bencentage of light is collected from a point source is as follows! NA =nsin(0/2) or, 0.45 = sin(0%) 07, (0/2) = 26,743 0 - 53.486 Y. of light collected = 2 ( cos(8/2)-1) x100 =-1 (0-893-1) x100 - % x 0.107 x 100 = 5.359. Hence, the percentage of light collected from a point source is 5.357. gf NA = I then, NA Ensino or a = sin(02) or o = of light collected = 2 (COR (2) - 1) x100 50 y. y

belo 168 voit In the case of water, Refractive Index (n) = 1,333. i en NA = sin( slumot at case-ll) I 96 NA=0.45 then out 0.45 = 1.38 Sino) or, sin(0/2) = 0.45 = 0.338 ( 1 + 33 or 0/2 2 19.755° (-Colo) cor, 0 39:51 x (recay of light collected = ( COD -1)x100 --(C09 (19.755)-1)x100 1x Folox = 19.15 y. lolos tigil 90 39 Cage -12), gG NA=1 then Triodo I = 1.33 sin (0/2) or, sin(x) = -0.751 ox; 0/2 = 48.672° 01, 0 = 97.354° y of light calkected = -1 [cog(9)-]X 100 201409) blac-[C08 (4867:)+] x 100 = 177.
t this is the solution of part(a) and part(b).