1. Refraction at a spherical interface between two (transparent) materials
All paraxial rays (making small angle with principal axis) from a point source will refract to the same point, the image.

2. Proof n1 sin 1 = n2 sin2, but for  <<1 (radian), sin   .
 paraxial rays: n1 1  n2 2 .

3. n1 sin 1 = n2 sin2, but for  <<1 (radian), sin   .
 paraxial rays: n1 1  n2 2 .
1 = +;  =  + 2
 n1 (+) = n2 ( - )
n1  + n2  = (n2 – n1) 

4. n1  + n2  = (n2 – n1) 
But   tan  = d/p
  tan  = d/q
  tan  = d/R
n1/p + n2/q = (n2 – n1)/R
Since this is true for all small angles, all paraxial rays from O are brought to focus at same point. QED

5. n1/p + n2/q = (n2 – n1)/R
Signs:
If the image is real (i.e. rays really converge), q > 0: This occurs if q is inside the second (n2) material, as above.
If transmitted rays are diverging, virtual image is formed inside the first (n1) material: q < 0.
R > 0 if convex surface faces object; R < 0 if concave surface faces object.
(opposite convention to mirrors)

6. Problem: n1 = 1 (air), n2 = 3/2 (glass), R = 5 cm
n1/p + n2/q = (n2 – n1)/R
 3/2q = 1/2R – 1/p
p = 20 cm: 3/2q = 1/10-1/20 = 1/20; q = 30 cm (real image inside glass)
p = 5 cm: 3/2q = 1/10 – 1/5 = -1/10; q = -15 cm (virtual image in the air)
p = 10 cm: 3/2q = 1/10 – 1/10 = 0; q =  (i.e. transmitted rays are all parallel)
p = : 3/2q = 1/2R = 1/10; q= + 15 cm

7. Problem: n1 = 3/2 (glass), n2 = 1 (air), R = 5 cm
n1/p + n2/q = (n2 – n1)/R
1/q = -1/2R – 3/2p < 0
 only virtual images
p = 20 cm: 1/q = -1/10 - 3/40 = -5/40; q = -8 cm (virtual image in the glass)
p = 5 cm: 1/q = -1/10 – 3/10 = -4/10; q = -2.5 cm (virtual image in the glass)
p = 10 cm: 1/q = 1/10 – 3/20 = -4/20; q = -5 cm (virtual image in the glass)
p = : 1/q =-1/2R = -1/10: q = -10 cm

8. Problem: n1 = 1 (air), n2 = 3/2 (glass), R = - 5 cm (concave toward object)
n1/p + n2/q = (n2 – n1)/R
 3/2q = 1/2R – 1/p < 0 (since R < 0): only virtual images
p = 20 cm: 3/2q = -1/10-1/20 = -3/20; q = cm (virtual image in the air)
p = 5 cm: 3/2q = -1/10 – 1/5 = -3/10; q = cm (virtual image in the air)
p = 10 cm: 3/2q = -1/10 – 1/10 = -1/5; q = -5cm (virtual image in the air)
p = : 3/2q = 1/2R = -1/10: q = -15 cm

9. If flat interface, R =  n1 n2
n1/p + n2/q = (n2 – n1)/R = 0: q = -(n2/n1) p < 0
only virtual images
n2 > n1
n2 < n1

10. Example: Looking at an object in the water
na = 1
q = -pna/nw = -3/4 p
(Fish in an aquarium look closer than they really are)
|q| p
nw = 4/3
(Refraction angle not drawn accurately, so figure not to scale)

11. Example: Observer in water looking at an object in air
Fly
|q|
na = 1
nw = 4/3
q = -pnw/na = -4/3 p
(Fly looks further away to the fish than it really is.)
(Refraction angle not drawn accurately, so figure not to scale)

12. R1 R2
I1 = O2 O1 n I2
If material has two refracting surfaces, it is a lens. The first surface forms an image which acts as an object for the second surface, which forms the final image.

13. Spherical Surfaces n1/p + n2/q = (n2-n1)/R
First surface: R1 >0, n2 = n, n1 = 1:
1/p1 + n/q1 = (n-1)/R1 1/q1 = (n-1)/R1 - 1/p1
[In this case, q1 <0 because (n-1)/R1 < 1/p1 and I1 is a virtual image, but acts as a “real object” for surface 2.]

14. Second surface: R2 < 0, n2 =1, n1 = n q2
n1/p + n2/q = (n2-n1)/R
Second surface: R2 < 0, n2 =1, n1 = n
n/p2 + 1/q2 = (1-n)/R2 = -(n-1)/R2 = (n-1)/|R2|
1/q2 = (n-1)/|R2| - n/p2
[q2>0 (real image) if (n-1)/|R2| > n / p2]

15. 1/p1 + 1/q2 = (n-1) (1/R1 – 1/R2) 1/p1 + n/q1 = (n-1)/R1 (1)|
n/p2 + 1/q2 = (1-n)/R2 = -(n-1)/R2
But p2 = t + |q1| = t – q1  -q1 if t small (thin lens approximation)
-n/q1 + 1/q2 = -(n-1)/R2 (2)
Add (1) + (2) to get net result:
1/p1 + 1/q2 = (n-1) (1/R1 – 1/R2)
Initial (actual) object
Final image

16. 1/f = (n-1) (1/R1 - 1/R2) (lens maker equation)
1/p1 + 1/q2 = (n-1) (1/R1 – 1/R2)
actual object
final image
1/p + 1/q = (n-1) (1/R1 – 1/R2)
Signs:
R’s are positive if convex side faces object
R’s are negative if concave side faces object
Real image if q > 0
Virtual image if q< 0
Magnification M  h’ / h = -q/p
If p = , 1/q = (n-1) (1/R1-1/R2)  1/f (focal point)
1/p + 1/q = 1/f
1/f = (n-1) (1/R1 - 1/R2)
(lens maker equation)

17. 1/p + 1/q = 1/f 1/f = (n -1) (1/R1 - 1/R2) True for other geometries
A lens which is fatter in center than edges will have f > 0 and is called a converging lens.
A lens which is thinner in center than edges will have f < 0 and is called a diverging lens.

18. Converging Lens 1/p + 1/q = 1/f >0
If p = , q = f: Light from infinity (i.e. parallel rays) will converge at the focal point on the opposite side of lens. (Real image)
Can also turn the arrows around: light from the focal point will emerge as parallel rays (i.e. be focused at infinity)
(i.e. if p = f, q = .)

19. Diverging Lens 1/p + 1/q = 1/f < 0
If p = , q = f = -|f|: Light from infinity (i.e. parallel rays) will diverge and appear to originate at the focal point on the front side of lens. (Virtual image)
Can also turn the arrows around: light that is heading toward the focal point will emerge as parallel rays (i.e. be focused at infinity)
(i.e. if p = f = - |f|, q = .)