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)}]-1
F = [ (1.458467-1) {(2/91.09) -(1.458467-1)/1.458467*3.78/91.092}]-1
F = 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-R
RR(n-1)/n*t = 100- (1.85-1)/1.85*2.5
R= 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 mm
Defocus = 3.59 mm

F/# is irrelevant. Included to challenge you!
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