Assignment 3, solutions:

Problem 1:

Optical path length between objective and eyepiece = f₁ + f₂.

f₁ = 10 m, magnitude of angular magnification = f₁/f₂ = 10, f₂ = 0.1 m.

f₁ + f₂ = 1.1 m.

Problem 2:

The transformation matrix from V to V’ is

.

With f₁ = 20 cm, f_i = 2 cm and f₂ = 5 cm we have 

.

This is not an image forming, but a telescopic system, since M₁₂ = 0.  M₁₁ = m_q = 4.

Problem 3:

The transformation matrix from V to V’ for two thin lenses in air, separated by a distance d, with focal lengths f₁ and f₂, respectively, has the form

.

If f₂ = f₁ and d = (3/4)f₁, then 

.

The effective focal length id f_eff = 4f₁/5.

The principal planes are located a distance D₁ = (1/M₁₂)(1-M₁₁) to the left of V and a distance D₂ = (1/M₁₂)(1-M₂₂) to the right of V'.

D₁ = D₂ = -3f₁/5, so the principal planes lie 3f₁/5 to the right of V and to the left of V', respectively.

The focal points lie 4f₁/5 to the left of D₁ and to the right of D₂, so they lie f₁/5 to the left of V and to the right of V', respectively.

Problem 4:

The transformation matrix from V to V’ is

.

Here V is the location of the mirror and V' is the location of the thin lens.

This is not an image forming, but a telescopic system, since M₁₂ = 0.  M₁₁ = m_q = -R/(2f_L).

Problem 5:

Image all optical elements into object and image space:

(a)  The aperture acts as the aperture stop.  It physically limits the solid angle of rays passing through the system from an on-axis object point.

(b)  The exit pupil is the image of the aperture stop as seen through all the optics beyond the aperture stop.  It can be a real or virtual image, depending on the location of the aperture stop.

To find its location we use 1/2 + 1/x_i = 1/f₂ = 1/3.  x_i = -6.  The exit pupil is a virtual image of the aperture stop.  Its location is 6 cm to the left of L₂.

M = -x_i/x_o = 6/2 = 3.  The size of the exit pupil is 3 times the size of the aperture stop.  Its diameter is 3 cm.

(c)  The entrance pupil is the image of the aperture stop in as seen through all the optics before the aperture stop.

To find its location we use 1/3 + 1/x_i = 1/f₁ = 1/9.  x_i = -4.5.  The entrance pupil is a virtual image of the aperture stop.  Its location is 4.5 cm to the right of L₁.

M = -x_i/x_o = 4.5/3 = 1.5.  The size of the entrance pupil is 1.5 times the size of the aperture stop.  Its diameter is 1.5 cm.

Problem 6:

Design of a Galilean telescope:

Use the convex lens with focal length +80 cm as the objective and the concave lens with focal length -20 cm as the eyepiece.  The distance between the lenses should be 60 cm.