Spherical Interface
A spherical interface is a surface between two regions with different refractive indices. For example, a rod polished so that its end is spherical is a type of interface. This section examines how light waves are changed as they pass through such an interface.
The derivation of the relevant equation proceeds as before using the diagram shown below. A ray OQ is refracted to QI according to Snell's Law. The normal to the surface is the line from the centre of curvature CQ.
As before:
Snell's Law becomes (in the paraxial approximation):
n1θ1=n2θ2
In triangles OQC and QCI
θ1 = α + β ...(2)
β = θ2 + γ = n1θ1 / n2 + γ ...(3)
From (2) and (3)
Together with (1)
Rearranging one can write:
This then takes the form
Vf = P + Vi
where
- Vf = ± n2/v is the final vergence
- Vi = ± n2/u is the initial vergence
- P = ± |n2 - n1|/r is the power of the surface
The sign convention is as before - the vergence is:
- + for converging waves
- - for diverging waves
Powers
An analysis of the equation for the power of a surface leads to the following easy to remember equation:
'In' refers to concave side, 'Out' is convex side. That is, they refer to the refractive index on the inside and outside of the curvature - NOT inside and outside the surface.
Focal Point
The focal points for an interface can be calculated in the same manner as that for a mirror. The first focal point is the position where a point source should be placed so that light passing through the interface will be collimated (Vf=0). The second focal point is where collimated light entering the system is focused (Vi=0). See if you can determine these locations.
Examples - Interfaces
Example 1. A cylindrical glass rod (n=1.5) with a hemispherical end of radius 25 mm is placed in air. What distance into the rod is an image formed of an object 100 mm away in air?
Step 1. Draw a diagram.
Step 2. Calculate the initial vergence: An object generates light rays that diverge. Therefore the initial vergence is negative and given by
Vi = ± n/u = -1/0.1 = -10 D
(The refractive index for air is approximately 1. Remember to convert all distances to metres!)
Step 3. Determine the power: Using the formula above, the rod has a hemispherical end (it curves out) so that the refractive index on the inside of the curvature is that of the glass.
P = (nin - nout) / r = (1.5 - 1) / 0.025 = +20 D
Step 4. Calculate the final vergence:
Vf = P +Vi = -10 + 20 D = +10 D
Step 5. Calculate the final distance. The vergence is positive so the rays are converging to a real image. Remember to use the refractive index of the glass as the light is now propagating through the rod.
Vf= ± n/v => v = n/Vf = 1.5 / 10 = 0.15 m = 150 mm.
Step 6. Compare this with your diagram. Do the numbers make sense??
Example 2. (Try this yourself) A cylindrical glass rod (n=1.5) with a hollowed hemispherical end of radius 25 mm is placed in air. Locate the position of an image formed of an object placed 100 mm away in air.
Step 1. Draw a diagram. This is the reverse case from above. The end of the rod is now concave.
Step 2. Calculate the initial vergence (it should be the same as before)
Step 3. Determine the power: Using the formula above, the rod has a hemispherical end that now curves inwards so that the refractive index on the inside of the curvature is that of the air.
Step 4. Calculate the final vergence:
Step 5. Calculate the final distance. The vergence should be negative so the rays are diverging forming a virtual image (on the same side as the object). Although the image is in the air, the rays forming that image are propagating in the glass so it is still necessary to use the refractive index of the glass rod.
Step 6. Compare this with your diagram. Do the numbers make sense??
You should find that a virtual image was formed 50 mm from the rod on the same side as the object.
More problems of a similar nature can be found in the on-line examples.