1. IMAGE FORMATION
e.g. lens, mirors
Spherical wavefronts expand, rays diverge
Spherical wavefronts contract, rays converge
The optical system may consist of reflecting and/or refracting surfaces that may change the direction of rays from an object point O.
Fermat’s principle implies that as every ray starts at O and ends at I, each ray takes the same transit time  isochronous rays.
Principle of reversibility states if I becomes the object point, each ray will reverse its direction but maintain its path and ends at its corresponding image point O  I and O are conjugate points of optical system.
- ideal optical system: every ray from object point O intercepted by the system must pass through its conjugate image point I

2. In non-ideal optical system, some rays leaving O may not reach I because of:
light scattering - reflection losses at refracting surfaces, diffuse reflections from reflecting surfaces, scattering by inhomogeneities in transparent media. These degrade image & cause loss of brightness.
aberrations - departure from ideal, paraxial imaging.
diffraction limited - effect due to using a limited portion of the wavefront being intercept by optical system.

3. Cartesian surfaces are surfaces that form perfect images.
Examples of Cartesian reflecting surfaces
Rays are isochronous.

4. Cartesian refracting surface:
Arbitrary point
Every ray from O that refracts at surface  is required to pass through image point I (e.g. isochronous rays OPI and OVI).
OPI and OVI are not equal in length because of different refractive indices of media on either side of , but have equal transit times.
Transit time of ray through medium of thickness x with refractive index n is:
v = speed in medium of refractive index n;
nx = optical path length (equivalent path of the ray if it were to travel in air at speed c)

5. constant (3.1) constant (3.2)
However, optical path of ray OPI = optical path of ray OVI
Thus,
constant
(3.1)
Rewriting in terms of (x,y) coordinates:
constant
(3.2)
Eq. (3.2) describes a Cartesian ovoid of revolution.

6. Hyperbolic surface images O at infinity when O is at one focus and ni > no
Ellipsoid surface images O at infinity when O is at one focus and no > ni
Cartesian refracting surfaces
Cartesian ovoid images O at I by refraction.

7. A lens refracts light rays twice, once at each surface, and can produce a real image outside the lens (image may be in same medium as object)
Double hyperbolic lens produces aberration-free image of point object O;
only if point O is at correct distance from lens and on axis
Shortcomings of hyperboloid surfaces:
difficult to fabricate
images of actual objects (extended) are not totally aberration-free
Therefore, most optical surfaces are spherical (ease in fabrication outweighs its spherical aberrations)

8. REFLECTION AT SPHERICAL SURFACE (with radius of curvature R)
Ray 1
Ray 2 R
(1) Spherical mirrors
Convex spherical mirror
(Center of curvature C on right of vertex V)
Rays 1 & 2 diverge after reflection.
Image point I conjugate to O is at intersection of rays 1 (extended backward) & 2 after reflection.
Image is VIRTUAL, located BEHIND the mirror surface.

9. To obtain a relationship between s and s’ in terms of R only:
s & s’ are respective object and image distances measured from vertex V.
From triangle POC:
(3.3)
(external angle = sum of two internal opposite angles)
From triangle POI:
(3.4)
(3.5)
Combining (3.3) & (3.4):
Using small-angle approximation (leads to first-order, or Gaussian, optics)
(3.6) 
(3.7) 

10. As paraxial optics is considered,  and  are small, axial distance VQ can be neglected, and angles , ’ and  in Eq.(3.5) may be replaced by their tangents as:
(3.8)
Canceling h:
(3.9)
Employing an appropriate sign convention to represent convex as well as concave mirrors, the mirror equation becomes:
(3.10)

11. Sign convention chosen is as follows (assume light ray to enter from the left):-
Object
Real
+ s
O is at left of V
Virtual
 s
O is at right of V
Image
+ s’
I is at left of V
 s’
I is at right of V
Radius of curvature
+ R when C is at right of V (CONVEX)
 R when C is at left of V (CONCAVE)
For plane mirror, R   and s’ = s (negative means virtual image for real object)

12.  +f f Object focus is defined by: and image focus is defined by:
Concave mirror
Convex mirror +f f
Object focus is defined by:
and image focus is defined by:
Therefore, 
(3.11)
Thus,
(3.12)
another form of mirror equation

13. To determine lateral magnification:
As i, r and  are equal,
(3.13)
Lateral magnification defined by ratio of lateral image size to corresponding object size:
(3.14)
Again employing the appropriate sign convention, lateral magnification is given by:
(3.15)

14. Object height
+ ho
Above axis, erect object
 ho
Below axis, inverted object
Image height
+ hi
 hi
Lateral magnification
+ m
Image has same orientation as object
 m
Image is inverted relative to object

15. Concave mirror Convex mirror Virtual, magnified, erect
Real, minified, inverted
Real, magnified, inverted
Virtual, magnified, erect
Always virtual, minified, erect
RO = real object
RI = real image
VI = virtual image
C = center of curvature
F = focal point
Image formation by spherical mirrors of object placed at different distances from the mirror surface.
Height of arrow from optical axis shows relative height of object/image.

16. Ray diagrams for spherical mirrors:
(a) Real image, concave mirror
(b) Virtual image, concave mirror
Intersection of any 2 rays will locate the image:
Ray-1: Line parallel to optical axis from tip of object to reflecting surface, then bends to pass through F after reflection
Ray-2: Line from tip of object to/away from F to reflecting surface before bending to become parallel to optical axis after reflection
Ray-3: Line from tip of object to/through reflecting surface to center of curvature.
(c) Virtual image, convex mirror

17. Example:
An object 4 cm high is placed 40 cm from (a) a convex, and (b) a concave spherical mirror, each of 20 cm focal length. Find the position and type of image in each case.
(a) Convex mirror: f =  20 cm and s = +40 cm
Using mirror equation:
cm (virtual, right of V)
(minified, erect)

18. (b) Concave mirror: f = +20 cm and s = +40 cm
Using mirror equation again:
cm (real, left of V)
(same size, inverted)

19. REFRACTION AT SPHERICAL SURFACE (with radius of curvature R)
Ray 2 refracted (Snell’s law)
Axial ray 1 refracted with no change
Triangle CPO:
(3.16)
Triangle CPI:
(3.17)
From Snell’s law:
(3.18)
For paraxial rays, we use approximation
Eq.(3.18) becomes
For small angles, distance QV can be neglected, and with approximation  
(3.19)
Employing the same sign convention as for mirrors, the refraction equation becomes:
(3.20)

20. Sign convention for spherical refracting surface:
Object
Real
+ s
O is at left of V
Virtual
 s
O is at right of V
Image
+ s’
I is at right of V
 s’
I is at left of V
Radius of curvature
+ R when C is at right of V (CONVEX)
 R when C is at left of V (CONCAVE)
For plane surface: R  , or
which is the apparent depth.

21. Lateral magnification of refracting surface is determined as:
From Snell’s law and small angle approximation: 
Lateral magnification:
(3.21)
(negative sign inserted according the sign convention used)

22. From refraction equation:
Example:
A real object in air (na = 1) is placed 30 cm from a convex spherical surface of radius R = +5 cm. On the right of the interface, the medium is water (nw = 4/3). Find the image distance s’ and lateral magnification m of the image.
From refraction equation:   cm
(real image, on right of refracting surface in water)
Magnification is
(inverted, same size)

23. Suppose now there exists a 2nd concave refracting surface that reduces the length of the 2nd medium to only 10 cm thick, forming a thick lens.
n3 = n1 = 1
The rays from object after refracting through 1st surface will still form a 1st image as calculated previously. But this image will now be the object for the 2nd surface, producing a different final image.
Real image of 1st surface is on right of 2nd surface  virtual object for 2nd surface at distance s2 = 30 cm

24. For 2nd surface:   
cm (real image, to right of 2nd surface)
Lateral magnification: (
the size of 1st image or 2nd object, i.e. size of original object;
same orientation as 1st image, i.e. inverted relative to original object)