### Optics Tutorial 7 HW solutions

 1.You have a microscope objective with an NA of 0.723. At what F/# is this operating? Compute the marginal ray angle incident on the microscope slide.  Solution: F/# = 1/2/NA = 1/2/0.723 = F/0.69 NA=sin(θ) or  θ = arcsin(NA) = arcsin(0.723) θ = 46.3° 2.A 100 mm focal length lens is used with a 2/3” detector. Compute the angular full field of view (corner to corner) for an infinite conjugate.  Solution: h = F tan(θHFOV) or θHFOV =  arctan[h/F] For a 2/3" sensor the image diagonal is 11mm (Reference) h is the semi-image diameter or 5.5 mm θFFOV =  2*θHFOV =  2*arctan[h/F] =  2*arctan[5.5 mm/ 100 mm]  θFFOV =  6.3° 3.An F/3 achromat is used as a 3:1 relay. Compute the working F/#.  Solution: F/#working = F/#infinite *(1-M) actually M = 3x for a simple lens since the image gets inverted. F/#working = F/3 *(1-[-3]) = F/3 *(4) F/#working = F/12 4.Compute the focal length to four significant figures for the Edmund Optics UV DCX lens 48297.  Solution: 1/F = (n-1) {(1/R1) –(1/R2)+(n-1)/n*t/R1/R2} F = [ (n-1) {(1/R1) –(1/R2)+(n-1)/n*t/R1/R2}]-1 n = 1.458467 (Corning C79-80) information not on the website... I found the lens in the Zemax library. Radius = 91.09 mm (biconvex) t = 3.78 mm Simplify for a biconvex F = [ (n-1) {(2/R) -(n-1)/n*t/R2)}]-1F = [ (1.458467-1) {(2/91.09) -(1.458467-1)/1.458467*3.78/91.092}]-1F = 99.9941 mm 5.If the front radius of a lens is 100 mm and its index is 1.85, what is the back radius to make is a zero power element if it is 2.5 mm thick?  Solution:R1-R2 = R2 = R1 - (n-1)/n*t = 100- (1.85-1)/1.85*2.5R2 = 98.85 mm 6.A 10 mm thick piece of N-BK7 is inserted into an F/1.0 converging beam. How far is focused moved? Repeat for an F/10 converging beam. Solution:Defocus = (n-1)/n*t   = (1.56-1)/1.56*10 mmDefocus = 3.59 mmF/# is irrelevant. Included to challenge you!
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