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 θ_{i} with respect to the normal, and the refracted angle
θ_{t} is given by Snell's law, n_{i}sinθ_{i }= n_{t}sinθ_{t}.

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 10^{o}.)

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

n_{1}/x_{o }+ n_{2}/x_{i }= (n_{2 }- n_{1})/R,

if we adopt the following sign conventions.

- The object's position x
_{o}is in front of the interface and the object distance is positive. - The image distance x
_{i}is positive if the image position lies behind the interface. - The image distance x
_{i}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 = h_{i}/h_{o }= -(n_{1}x_{i})/(n_{2}x_{o}).
If M is negative, the image is inverted.

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 x_{o} is very large and n_{1}/x_{o}
is very small.

We can safely ignore this term in the equation n_{1}/x_{o
}+ n_{2}/x_{i }= (n_{2 }- n_{1})/R.

We then have n_{2}/x_{i }= (n_{2 }- n_{1})/R,

which yields x_{i }= n_{2}R/(n_{2 }- n_{1}) =
(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 n_{1}/x_{o }+ n_{2}/x_{i }= 0.

Let an object be underwater, 10 cm from the water-air interface.

We find x_{i }= -(n_{2}/n_{1})x_{o }= -(1/1.33)x_{o}. x_{i}
= -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.

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:

x_{o }= 50 cm. x_{i
}= -(1/1.309)x_{o
}= -38.2 cm.

The
apparent position is 38.2 cm below the top surface of the ice.