Assignment 1, solutions:

Problem 1:

For a thick lens we have for the transformation between the vertices

with P = (n'-n₁)/R₁ and P' = -(n'-n₂)/R₂.  For our lens P = P' = -1/100

The matrix for the transformation between the principal planes is 

,

with P_syst = -M₁₂.

For a thick lens we therefore have P_syst = (n'-n₁)/R₁ - (n'-n₂)/R₂ + D(n'-n₁)(n'-n₂)/(n'R₁R₂).

Given the location of the principal planes we have 

n₁/S₁ + n₂/S₂ = P_syst,

f' = n₂/P_syst,  f = n₁/P_syst.

In our problem n₁ = n₂ = 1,  n' = 1.5,  R₁ = -50 mm, R₂ = +50 mm, D = 15 mm.  We have f' = f = 1/P_syst, or

1/f = (n'-1)(1/R₁ - 1/R₂) + D(n'-1)²/(n'R₁R₂),

1/f = (-0.02 - 15(0.5)²/3750) mm^-1 = -0.021 mm^-1,   f = -47.619 mm.

Here f is the distance from the principal planes to the focal points.  To find the distance from the vertex V to the focal point F we have to add D₁ = n₁P'D/(-P_systn’) to f.  To find the distance from the vertex V' to the focal point F' we have to add D₂ = n₂PD/(-P_systn’) to f'.

D₁ = f/10 = -4.7619 mm,   VF = -52,38 mm

D₂ = f/10 = -4.7619 mm,   V'F' = -52,38 mm

F is 52,38 mm to the right of V, F ' is 52,38 mm to the left of V '.

Problem 2:

surface data:

lens drawing:

Problem 3:

surface data:

lens drawing:

Problem 4:

1/x_o + 1/x_i = 1/f.   x_i = 2x_o/(x_o-2)

Virtual images:

If  x₀ < 2 in then x_i is negative.  We have a virtual image. The rays are diverging as they enter the tube.  The location of the aperture stop is always at 5 in at the end of the tube.

Real images:

If x_i = 2x_o/(x_o-2) < 2.5 in or  x_o > 10 in then the aperture stop is the section of the tube a distance 2x_i to the right of the lens.  It physically limits the solid angle of rays passing through the system from an on-axis object point.  This is the case as long as the image distance is less than 2.5 in, or twice the image distance is less than the length of the tube.

The lens is the aperture stop is the image distance is positive and larger than 2.5 in.  If 2 < x_o < 10 then the aperture stop is the lens.