Assignment 2, solutions:

Problem 1:

Let the initial and final refracting surfaces intersect the optical axis at V and V’.  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

,

det(M) = 1.

The effective focal length if f_eff, where 1/f_eff = P_sys = -M₁₂.

1/f_eff =1/f₁ + 1/f₂ – d/(f₁f₂).

Problem 2:

1/f_eff = 1/f₁ + 1/f₂ – d/(f₁f₂)

The focal lengths f₁ and f₂ depend on n.  For a thin lens

1/f = (n-1)(1/R₁ - 1/R₂) = (n-1)A, where A is a constant.

Therefore 

1/f_eff = (n-1)A₁ + (n-1)A₂ - (n-1)²A₁A₂d.

Let d(1/f_eff)/dn = A₁ + A₂ - 2(n-1)A₁A₂d = 0.  Then first order variations of the effective focal length with n are zero.  Then 

1/((n-1)f₁) + 1/((n-1)f₂) – 2d/((n-1)f₁f₂) = 0,

d = (1/2)(f₁ + f₂).

Problem 3:
O and I are conjugate planes.  To find the radius of curvature we use 

.

1/24 + 1.5/36 = 0.5/R.  R = 6 units.
Now we can find the matrix connecting the planes.

The lateral magnification is m_x = -1 and the ray angle magnification is m_q = -1/1.5.

Problem 4:

See the spreadsheet.  To make an approximately plane surface, choose a very large R.

For example:

Problem 5:

.

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

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'.  (Here we use the convention that after reflection our rays continue to travel to the right.)

D₁ = 100, so the first principal planes lies 100 cm to the left of V.

D₂ = 80, so the second principal planes lies 80 cm to the right of V' (i.e. in front of the mirror).

1/f_eff = -M₁₂, f_eff = - 40 cm, so the focal points lie 40 cm to the right of the first principal plane and 40 cm to the left of the second principal plane.  (That means 60 cm in front of the lens and 40 cm in front of the mirror, respectively.)

We have 1/S₁ + 1/S₂ = 1/f_eff,  S₁ = -85 cm, the object lies 85 cm to the right of the first principal plane.

S₂ = -75.556 cm, the image lies (80 - 75.556) cm = 4.44 cm in front of the mirror surface after reflection.

Here V and V' are at the location of the lens, before and after passage through the system, respectively.

D₁ = 20, so the first principal planes lies 20 cm to the left of V.

D₂ = 20, so the second principal planes lies 20 cm to the right of V' (i.e. in front of the lens after reflection). 

1/f_eff = -M₁₂, f_eff = - 40/3 cm, so the focal points lie 40/3 cm to the right of the first principal plane and 40/3 cm to the left of the second principal plane. (That means the focal points lie 20/3 cm in front of the lens.)

We have 1/S₁ + 1/S₂ = 1/f_eff,  S₁ = -5 cm, the object lies 5 cm to the right of the first principal plane.

S₂ = 8 cm, the image lies (20 + 8) cm = 28 cm in front of the lens after reflection.

Problem 6:

The transformation from the front to the back of the sphere yields

.

For incoming rays parallel to the optic axis q = 0.  They form an image at x₁ = 0 on the far surface of the sphere.

x₁ = 0 at z = 2R for any x yields

.

This equation yields n = 2.