## A single interface

Images can be formed by reflection and by refraction.  Assume that we have a boundary between two media with different indices of refraction, and that the boundary has a radius of curvature R, as shown in the figure.  The center of curvature is at the origin.
Light is refracted at the boundary.  Two rays starting at an object are shown.  One ray is incident normally and is not refracted.  The other ray is incident at an angle θi with respect to the normal, and the refracted angle θt is given by Snell's law, nisinθi = ntsinθt.
An image is formed where the two rays intersect.

For small angles θ, (θ(rad) << 1) we have sinθ ≈ tanθ ≈ θ.
(See the graph on the right which plots θ(rad), sinθ, and tanθ versus θ(deg) for angles from 0 to 10o.)
This is called the small angle approximation.  It simplifies calculations if it is applicable.

Using the small angle or paraxial approximation we can show that for a spherical interface

n1/xo + n2/xi = (n2 - n1)/R,

if we adopt the following sign conventions.

• The object's position xo is in front of the interface and the object distance is positive.
• The image distance xi is positive if the image position lies behind the interface.
• The image distance xi is negative if the image position lies in front of the interface.
• The radius of curvature R is positive, if the center of curvature lies behind the interface.
• The radius of curvature R is negative, if the center of curvature lies in front of the interface.

The magnification is M = hi/ho = -(n1xi)/(n2xo).  If M is negative, the image is inverted.

#### Problem:

A simple model of the human eye ignores its lens entirely.  Much of what the eye does to light happens at the transparent cornea.  Assume that this outer surface has a radius of curvature of 6 mm, and assume that the eyeball contains just one fluid of index of refraction n = 1.4.  Prove that every distant object will be imaged on the retina, 21 mm behind the cornea.  Describe the image.

Solution:
For distant objects xo is very large and n1/xo is very small.
We can safely ignore this term in the equation n1/xo + n2/xi = (n2 - n1)/R.
We then have n2/xi = (n2 - n1)/R,
which yields xi = n2R/(n2 - n1) = (1.4*6 mm)/0.4 = 21 mm.
The image distance is 21 mm.
The image is very small and inverted.

For a flat interface we have R = infinite and n1/xo + n2/xi = 0.
Let an object be underwater, 10 cm from the water-air interface.
We find xi = -(n2/n1)xo = -(1/1.33)xo.  xi = -7.5 cm.
The magnification is M = 1, it is positive.
The image is a virtual, non-inverted image.
A detector viewing the object from the air side sees the image at the apparent position closer to the interface.

#### Problem:

A cubical block of ice, 50 cm on a side, is placed on a level floor over a speck of dust.  Find the location of the image of the dust if the index of refraction of the ice is n = 1.309.

Solution:
xo = 50 cm.  xi = -(1/1.309)xo = -38.2 cm.
The apparent position is 38.2 cm below the top surface of the ice.