1. Mirrors continued

2. We will continue with mirrors by looking at convex or diverging mirrors.
Diverging mirrors differ from converging mirrors by having the focal point and the centre of curvature behind the mirror. They are virtual.
The 3 rays that we used to locate the image with converging mirrors are the same for diverging mirrors.
The ray parallel to the principal axis is reflected back so it appears to be coming from the focal point.
The ray that is directed to the focal point is reflected back parallel to the principal axis.
The ray that is directed toward the centre of curvature is reflected back along the same path.

3. Diverging (convex) mirror

4. It is important to remember that regardless of where the object is located in front of a diverging (convex) mirror, the characteristics of all images are:
ERECT
REDUCED IN SIZE
VIRTUAL
LOCATED BEHIND THE MIRROR BETWEEN THE VERTEX AND THE FOCAL POINT.
Remember the images for converging mirrors are:
INVERTED
REDUCED (UNLESS THE OBJECT IS BETWEEN F AND C)
REAL
LOCATED IN FRONT OF THE MIRROR EXCEPT WHEN THE OBJECT IS LOCATED BETWEEN THE VERTEX AND FOCAL POINT. THEN THE IMAGE IS ERECT, LARGER, AND VIRTUAL.

5. OPTICS
Up to now we have just looked at mirrors. We now will discuss the refraction of light and how light behaves passing through lenses.
When light travels from air into another medium, like water, its path is bent as it crosses the boundary between the 2 mediums.
This bending of the beam is called refraction.
The amount of refraction depends on the properties of the 2 media and the angle at which the light strikes the boundary.
Partial reflection will occur when some of the light is reflected at the boundary layer, and partial refraction will happen when some of the light is transmitted through the boundary layer from one medium into the other medium.
When this happens, the direction of the light rays change.

6. Incident ray
Reflected ray
air
water
Refracted ray

7. Terms used when talking about lenses
Incident ray – the ray approaching the boundary
Reflected ray – the ray reflected at the boundary
Point of incidence – the point where the ray strikes the boundary
Normal – the imaginary line drawn at 90◦ to the boundary at the point of incidence.
Angle of incidence ( I ) – the angle between the incident ray and normal.
Angle of reflection ( r )– the angle between the reflected ray and normal.
Angle of refraction ( R )– the angle between the refracted ray and normal.

8. Snell’s law of refraction
Snell’s law can be summed up as:
When travelling from a medium of low index of refraction to a medium of high index of refraction, light bends towards the normal.
When travelling from a medium of high index of refraction to a medium of low index of refraction, the light bends away from the normal.
If the light ray travels along the normal to the boundary between the two media (angle of incidence is 0◦), then the light is not refracted as it enters the second medium (angle of refraction is 0◦).

9. the ratio of sin I over sin R sin I sin R
Snell’s law can also be summed up mathematically :
the ratio of sin I over sin R sin I
sin R
is constant for a specific colour of light and for two given materials.
The incident ray and the refracted ray are on opposite sides of normal
The incident ray, the refracted ray, and the normal all lie in the same plane.
N1 sin I = N2 sin R
(N1 and N2 are the indices of refraction for medium 1 and medium 2)

10. examples Note: The N for air is 1.0003 (1.0003)(sin 37° ) = (1.36)(x)
A laser beam is incident upon ethanol (N = 1.36) at an angle of incidence of 37.0 degrees. What is the angle of refraction?
solution: use the formula N1 sin I = N2 sin R
Note: The N for air is
(1.0003)(sin 37° ) = (1.36)(x)
(x) = (1.0003)(sin 37° ) ÷ (1.36)
= ÷ =
take sin-1 of to get 26.3 °

11. (1.0003)(sin 30.0°) = (1.52)(X) (x) = (1.0003)(sin 30.0°) ÷ (1.52)
A light beam in air hits a sheet of crown glass (N = 1.52) at an angle of 30.0 °.
What is the angle of refraction of the light beam?
N1 sin I = N2 sin R
(1.0003)(sin 30.0°) = (1.52)(X)
(x) = (1.0003)(sin 30.0°) ÷ (1.52)
= / 1.52 =
= sin = 19.2 °

12. Total internal reflection
Total internal reflection is an interesting phenomenon that occurs when light travelling through an area with a higher index of refraction to an area with a lower index of refraction hits a boundary that exceeds the critical angle and all the light reflects back into the area with the higher index of refraction.
The critical angle is : c = NR / Ni = index of refraction of the refractive medium divided by the index of refraction of the incident medium.

13. What is the critical angle in flint glass when light passes from flint glass to air?
Flint glass N = 1.62
c = NR / Ni
= / 1.62
= sin
= 38.1 °
Note: For questions where you are given the critical angle and one of the indices of refraction (with you expected to find the unknown index of refraction), you must take the sin of the critical angle.
Ex. Given the critical angle of 28 ° and the refractive index of air (1.0003), find the other index of refraction.
Ans. c = NR / Ni  sin 28 ° = / x  x = / = 2.13