Main points

Geometrical optics is the model we use if the wavelength of the light is very small compared to the dimensions of the equipment used to set up an optical system, and if the the photon energies are much smaller than the energy sensitivity of the equipment.  Then light can be modeled as a collection of rays.

Geometrical optics predicts that in a medium with constant index of refraction a ray's path is a straight line.  At an interface of two media with different indices of refraction n and n', the ray obeys Snell’s law, n sinθ  = n' sinθ'.  When reflecting from a surface, the ray obeys the law of reflection, θ_incident = θ_reflected.

Often we use geometrical optics to study image formation by reflection or refraction at spherical interfaces.  If we restrict ourselves to rays that stay close to the optical axis of an optical system, we can study image formation using the paraxial approximation.  In this approximation spherical surfaces can form perfect images.  In the paraxial approximation a ray's behavior at each spherical interface and its propagation between interfaces is described by a 2 by 2 matrix, and the ray is traced through the optical system by multiplying these 2 by 2 matrices.

If the paraxial approximation is not valid, and the ray does not stay close to the optical axis, then spherical interfaces do not form perfect images.  The image of a point is no longer a point, but a blob.   The shape of the blob depends on the object points position and on where exactly we put the image plane. We can define various types of aberrations.

Examples (Paraxial approximation)

Problem:

A thick biconvex lens in air has a radius of curvature of 10 cm at each surface and the surfaces are 10 cm apart.  Find the focal points and the focal lengths of the lens if the index of the glass of the lens is 1.50.

Solution:

• Reasoning:
  For a thick lens we have for the transformation between the vertices
  
  with P = (n' - n₁)/R₁ and P' = -(n' - n₂)/R₂.
  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.
  Here f' is the image focal length, when the object distance goes to -∞, f is the object focal length, when the image distance goes to ∞.
   

• Details of the calculation:
  In our problem n₁ = n₂ = 1,  n' = 1.5, R₁ = 10 cm, R₂ = -10 cm, D = 10 cm.
  We have f' = f = 1/P_syst, or
  1/f = (n' - 1)(1/R₁ - 1/R₂) + D(n' - 1)²/(n'R₁R₂),
  1/f = (0.5/5 - 10(0.5)²/150) cm^-1,  f = 12 cm.
  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 = -0.5f/1.5 = -(1/3)f,  VF = 2/3f = 8 cm.
  D₂ = -0.5f'/1.5 = -(1/3)f,  V'F' = 2/3f' = 8 cm.
  D₁ and D₂ are negative, each principal plane lies 4 cm inward from its vertex.
  F is 8 cm to the left of V, F ' is 8 cm to the right of V '.

Problem:

Let the lens be a biconcave thick lens with radii of 10 cm, index 1.5, thickness 1 cm.  Show that f = f'= -9.84 cm, VF = -10.16 cm, V 'F ' = -10.16 cm.

Solution:

• Reasoning:
  See the previous problem.
• Details of the calculation:
  In this problem n₁ = n₂ = 1,  n' = 1.5, R₁ = -10 cm, R₂ = 10 cm, D = 1 cm. 
  We have f' = f= 1/P_syst, or
  1/f = (n' - 1)(1/R₁ - 1/R₂) + D(n' - 1)²/(n'R₁R₂),
  1/f = (-0.5/5 - (0.5)²/150) cm^-1,   f = -9.84 cm.
  D₁ = P'D/(-P_systn’),  P'= -(n' - 1)/R₂,  D₁ = 0.5f/15,  VF = 1.033f = -10.16cm.
  The object focal plane lies to the right of V.
  D₂ = PD/(-P_systn’),  P= (n' - 1)/R₂,  D₂ = 0.5f'/15,  V'F' = 1.033f' = -10.16cm.
  The image focal plane lies to the left of V.

Problem:

A biconvex lens of index 1.50 has radii 10 cm and 20 cm, a thickness of 10 cm.  Air is to the left of V and water (of index 4/3) is to the right of V '.  Show that f' = 24 cm, f = 18 cm, V 'F ' = 16 cm, VF = 17 cm.

Solution:

• Reasoning:
  See the previous problem.
• Details of the calculation:
  In this problem n₁ = 1, n₂ = 4/3,  n' = 1.5, R₁ = 10 cm, R₂ = -20 cm, D = 10 cm. 
  P_syst = (n' - n₁)/R₁ - (n' - n₂)/R₂ + D(n' - n₁)(n' - n₂)/(n'R₁R₂),
  P_syst = (0.05 + 0.00833  - 10(0.5)(0.01667)/300) cm^-1 = 0.0556 cm^-1.
  f' = n₂/P_syst = 4/(3P_syst) = 24 cm,  f = n₁/P_syst = 1/P_syst = 18 cm.
  D₁ = n₁ P'd/(-P_systn’), P' = -(n'-n₂)/R₂ = 0.00833 cm^-1, D₁ = -1 cm, VF = 17 cm.
  D₂ = n₂Pd/(-P_systn’), P = (n'-n₁)/R₁ = 0.05 cm^-1, D₂ = -8cm, VF = 16 cm.